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Quadratic_Equations/113899: Hi
I am having great difficulty trying to solve the following:-
I have tried puting the equation for the first part of the problem into my TI-83 graphing calculator..but I cannot find a proper solution .. in fact I have tried all I can think of, this problem is causing me to fall behind with my work.
I have seen many examples of how to solve the problem for a rocket ..the fact that this is a two stage rocket..is causing me to not be able to solve it


I have spent ages trying to work out the way to program my TI-83 calculator to solve the following... I would appreciate guidance or direction to a site that can show me how to solve the following;-
the motion of two stage rocket is stated by it height y in metres above the ground since t seconds since launch.
Flies vertically Ist stage burns for 5 seconds and cuts out. 2nd stage rocket is under gravity alone.
modelled by the equations
Stage 1 height y = 10t2
time 0< t <5
stage 2 height y = - 5t2 + 150t - 375
time t >5
Question asks you plot function on ti 83, using an appropriate window for time(s)
hence sketch a position time graph for each stage of flight adding labels etc..
also ask you to plot use the n deriv function to plot rate of change of function.
If I can find the basic means of plotting graph I should be able to answer the question but cannot find anything in manual that is similar to solving a question of this type.
I appreciate yout time and effort and thanks !!!!!!!!
1 solutions

Answer 82894 by jim_thompson5910(28598) About Me  on 2007-12-02 12:50:18 (Show Source):
You can put this solution on YOUR website!
In order to graph pieces of a graph simply type


y1=(10x^2)(0< x <5)

and
y2=(5x^2 + 150x - 375)(x>5)


Notice how the interval is right next to the function. This is how you enter piecewise functions into a TI calculator




Graphs/113911: Find an equation for the line with y-intercept 3 that is perpendicular to the line: y=2/3x-4
1 solutions

Answer 82892 by jim_thompson5910(28598) About Me  on 2007-12-02 12:44:16 (Show Source):
You can put this solution on YOUR website!
Remember if a line has the y-intercept 3, then it goes through the point (0,3)

Solved by pluggable solver: Finding the Equation of a Line Parallel or Perpendicular to a Given Line


Remember, any two perpendicular lines are negative reciprocals of each other. So if you're given the slope of 2%2F3, you can find the perpendicular slope by this formula:

m%5Bp%5D=-1%2Fm where m%5Bp%5D is the perpendicular slope


m%5Bp%5D=-1%2F%282%2F3%29 So plug in the given slope to find the perpendicular slope



m%5Bp%5D=%28-1%2F1%29%283%2F2%29 When you divide fractions, you multiply the first fraction (which is really 1%2F1) by the reciprocal of the second



m%5Bp%5D=-3%2F2 Multiply the fractions.


So the perpendicular slope is -3%2F2



So now we know the slope of the unknown line is -3%2F2 (its the negative reciprocal of 2%2F3 from the line y=%282%2F3%29%2Ax-4). Also since the unknown line goes through (0,3), we can find the equation by plugging in this info into the point-slope formula

Point-Slope Formula:

y-y%5B1%5D=m%28x-x%5B1%5D%29 where m is the slope and (x%5B1%5D,y%5B1%5D) is the given point



y-3=%28-3%2F2%29%2A%28x-0%29 Plug in m=-3%2F2, x%5B1%5D=0, and y%5B1%5D=3



y-3=%28-3%2F2%29%2Ax%2B%283%2F2%29%280%29 Distribute -3%2F2



y-3=%28-3%2F2%29%2Ax-0%2F2 Multiply



y=%28-3%2F2%29%2Ax-0%2F2%2B3Add 3 to both sides to isolate y

y=%28-3%2F2%29%2Ax-0%2F2%2B6%2F2 Make into equivalent fractions with equal denominators



y=%28-3%2F2%29%2Ax%2B6%2F2 Combine the fractions



y=%28-3%2F2%29%2Ax%2B3 Reduce any fractions

So the equation of the line that is perpendicular to y=%282%2F3%29%2Ax-4 and goes through (0,3) is y=%28-3%2F2%29%2Ax%2B3


So here are the graphs of the equations y=%282%2F3%29%2Ax-4 and y=%28-3%2F2%29%2Ax%2B3




graph of the given equation y=%282%2F3%29%2Ax-4 (red) and graph of the line y=%28-3%2F2%29%2Ax%2B3(green) that is perpendicular to the given graph and goes through (0,3)




Inequalities/113912: x>-2
1 solutions

Answer 82891 by jim_thompson5910(28598) About Me  on 2007-12-02 12:41:56 (Show Source):
You can put this solution on YOUR website!
Do you want to graph?


Start with the given inequality:

x%3E-2

Set up a number line:
number_line%28500%2C-12%2C8%29

Now plot the point x=-2 on the number line


number_line%28500%2C-12%2C8%2C+-2%29


Now pick any test point you want, I'm going to choose x=0, and test the inequality x%3E-2


0%3E-2 Plug in x=0


Since this inequality is true, we simply shade the entire portion in which contains the point x=0 using the point x=-2 as the boundary.This means we shade everything to the right of the point x=-2 like this:
Graph of x%3E-2 with the shaded region in blue
note: at the point x=-2, there is an open circle. This means the point x=-2 is excluded from the solution set.



Equations/113909: 4x-40=8
1 solutions

Answer 82888 by jim_thompson5910(28598) About Me  on 2007-12-02 12:25:04 (Show Source):
You can put this solution on YOUR website!

4x-40=8 Start with the given equation



4x=8%2B40Add 40 to both sides


4x=48 Combine like terms on the right side


x=%2848%29%2F%284%29 Divide both sides by 4 to isolate x



x=12 Divide

--------------------------------------------------------------
Answer:
So our answer is x=12


Equations/113908: 7x=84
1 solutions

Answer 82887 by jim_thompson5910(28598) About Me  on 2007-12-02 12:21:52 (Show Source):
You can put this solution on YOUR website!

7x=84 Start with the given equation



x=%2884%29%2F%287%29 Divide both sides by 7 to isolate x



x=12 Divide

--------------------------------------------------------------
Answer:
So our answer is x=12


Equations/113907: x+58=67
1 solutions

Answer 82886 by jim_thompson5910(28598) About Me  on 2007-12-02 12:17:17 (Show Source):
You can put this solution on YOUR website!

x%2B58=67 Start with the given equation



x=67-58Subtract 58 from both sides


x=9 Combine like terms on the right side

--------------------------------------------------------------
Answer:
So our answer is x=9



Complex_Numbers/113855: Combine the complex numbers: (7-i) + (-3 + 4i).
1 solutions

Answer 82885 by jim_thompson5910(28598) About Me  on 2007-12-02 12:11:25 (Show Source):
You can put this solution on YOUR website!
Start with the given expression

Group like terms (i.e. pair the real components with each other and pair the imaginary components with each other)

Combine like terms. So it is now in a%2Bbi form where a=4 and b=3


Complex_Numbers/113856: Factor: 12a^2 + 5a - 2.
1 solutions

Answer 82884 by jim_thompson5910(28598) About Me  on 2007-12-02 12:10:22 (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)
In order to factor 12%2Aa%5E2%2B5%2Aa-2, first multiply the leading coefficient 12 and the last term -2 to get -24. Now we need to ask ourselves: What two numbers multiply to -24 and add to 5? Lets find out by listing all of the possible factors of -24


Factors:

1,2,3,4,6,8,12,24,

-1,-2,-3,-4,-6,-8,-12,-24, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to -24.

(-1)*(24)=-24

(-2)*(12)=-24

(-3)*(8)=-24

(-4)*(6)=-24

Now which of these pairs add to 5? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 5

||||||||
First Number|Second Number|Sum
1|-24|1+(-24)=-23
2|-12|2+(-12)=-10
3|-8|3+(-8)=-5
4|-6|4+(-6)=-2
-1|24|(-1)+24=23
-2|12|(-2)+12=10
-3|8|(-3)+8=5
-4|6|(-4)+6=2


We can see from the table that -3 and 8 add to 5. So the two numbers that multiply to -24 and add to 5 are: -3 and 8

So the original quadratic


12%2Aa%5E2%2B5%2Aa-2


breaks down to this (just replace 5%2Aa with the two numbers that multiply to -24 and add to 5, which are: -3 and 8)


12%2Aa%5E2%2Bhighlight%28-3a%2B8a%29-2 Replace 5%2Aa with -3a%2B8a

Group the first two terms together and the last two terms together like this:

%2812%2Aa%5E2-3a%29%2B%288a-2%29

Factor a 3a out of the first group and factor a 2 out of the second group.


3a%284a-1%29%2B2%284a-1%29


Now since we have a common term 4a-1 we can combine the two terms.


%283a%2B2%29%284a-1%29 Combine like terms.
==============================================================================

Answer:


So the quadratic 12%2Aa%5E2%2B5%2Aa-2 factors to %283a%2B2%29%284a-1%29




Notice how %283a%2B2%29%284a-1%29 foils back to our original problem 12%2Aa%5E2%2B5%2Aa-2. This verifies our answer.


Complex_Numbers/113857: Factor completely: x^2 - 8x + 16.
1 solutions

Answer 82883 by jim_thompson5910(28598) About Me  on 2007-12-02 12:08:23 (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)
In order to factor x%5E2-8%2Ax%2B16, first multiply the leading coefficient 1 and the last term 16 to get 16. Now we need to ask ourselves: What two numbers multiply to 16 and add to -8? Lets find out by listing all of the possible factors of 16


Factors:

1,2,4,8,16,

-1,-2,-4,-8,-16, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to 16.

1*16=16

2*8=16

4*4=16

(-1)*(-16)=16

(-2)*(-8)=16

(-4)*(-4)=16

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to -8? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -8

||||||
First Number|Second Number|Sum
1|16|1+16=17
2|8|2+8=10
4|4|4+4=8
-1|-16|-1+(-16)=-17
-2|-8|-2+(-8)=-10
-4|-4|-4+(-4)=-8


We can see from the table that -4 and -4 add to -8. So the two numbers that multiply to 16 and add to -8 are: -4 and -4

So the original quadratic


x%5E2-8%2Ax%2B16


breaks down to this (just replace -8%2Ax with the two numbers that multiply to 16 and add to -8, which are: -4 and -4)


x%5E2%2Bhighlight%28-4x-4x%29%2B16 Replace -8%2Ax with -4x-4x

Group the first two terms together and the last two terms together like this:

%28x%5E2-4x%29%2B%28-4x%2B16%29

Factor a 1x out of the first group and factor a -4 out of the second group.


1x%28x-4%29%2B-4%28x-4%29


Now since we have a common term x-4 we can combine the two terms.


%28x-4%29%28x-4%29 Combine like terms.
==============================================================================

Answer:


So the quadratic x%5E2-8%2Ax%2B16 factors to %28x-4%29%28x-4%29

which can also be written as %28x-4%29%5E2 since the factors repeat themselves


Notice how %28x-4%29%28x-4%29 foils back to our original problem x%5E2-8%2Ax%2B16. This verifies our answer.


Rational-functions/113863: My question pertains to solving linear equations using subsitution method.
Solve:
5x+2y = 0
x-3y = 0
Please show me detailed steps on doing a problem of this type - Thank you kindly!
1 solutions

Answer 82882 by jim_thompson5910(28598) About Me  on 2007-12-02 12:07:20 (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Solving a linear system of equations by subsitution


Lets start with the given system of linear equations

5%2Ax%2B2%2Ay=0
1%2Ax-3%2Ay=0

Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to choose y.

Solve for y for the first equation

2%2Ay=0-5%2AxSubtract 5%2Ax from both sides

y=%280-5%2Ax%29%2F2 Divide both sides by 2.


Which breaks down and reduces to



y=0-%285%2F2%29%2Ax Now we've fully isolated y

Since y equals 0-%285%2F2%29%2Ax we can substitute the expression 0-%285%2F2%29%2Ax into y of the 2nd equation. This will eliminate y so we can solve for x.


1%2Ax%2B-3%2Ahighlight%28%280-%285%2F2%29%2Ax%29%29=0 Replace y with 0-%285%2F2%29%2Ax. Since this eliminates y, we can now solve for x.

1%2Ax-3%2A%280%29-3%28-5%2F2%29x=0 Distribute -3 to 0-%285%2F2%29%2Ax

1%2Ax%2B0%2B%2815%2F2%29%2Ax=0 Multiply



1%2Ax%2B0%2B%2815%2F2%29%2Ax=0 Reduce any fractions

1%2Ax%2B%2815%2F2%29%2Ax=0%2B0Add 0 to both sides


1%2Ax%2B%2815%2F2%29%2Ax=0 Combine the terms on the right side



%282%2F2%29%2Ax%2B%2815%2F2%29x=0 Make 1 into a fraction with a denominator of 2

%2817%2F2%29%2Ax=0 Now combine the terms on the left side.


cross%28%282%2F17%29%2817%2F2%29%29x=%280%2F1%29%282%2F17%29 Multiply both sides by 2%2F17. This will cancel out 17%2F2 and isolate x

So when we multiply 0%2F1 and 2%2F17 (and simplify) we get



x=0 <---------------------------------One answer

Now that we know that x=0, lets substitute that in for x to solve for y

1%280%29-3%2Ay=0 Plug in x=0 into the 2nd equation

0-3%2Ay=0 Multiply

-3%2Ay=0%2B0Add 0 to both sides

-3%2Ay=0 Combine the terms on the right side

cross%28%281%2F-3%29%28-3%29%29%2Ay=%280%2F1%29%281%2F-3%29 Multiply both sides by 1%2F-3. This will cancel out -3 on the left side.

y=0%2F-3 Multiply the terms on the right side


y=%2B0 Reduce


So this is the other answer


y=%2B0<---------------------------------Other answer


So our solution is

x=0 and y=%2B0

which can also look like

(0,%2B0)

Notice if we graph the equations (if you need help with graphing, check out this solver)

5%2Ax%2B2%2Ay=0
1%2Ax-3%2Ay=0

we get


graph of 5%2Ax%2B2%2Ay=0 (red) and 1%2Ax-3%2Ay=0 (green) (hint: you may have to solve for y to graph these) intersecting at the blue circle.


and we can see that the two equations intersect at (0,%2B0). This verifies our answer.


-----------------------------------------------------------------------------------------------
Check:

Plug in (0,%2B0) into the system of equations


Let x=0 and y=%2B0. Now plug those values into the equation 5%2Ax%2B2%2Ay=0

5%2A%280%29%2B2%2A%28%2B0%29=0 Plug in x=0 and y=%2B0


0%2B0=0 Multiply


0=0 Add


0=0 Reduce. Since this equation is true the solution works.


So the solution (0,%2B0) satisfies 5%2Ax%2B2%2Ay=0



Let x=0 and y=%2B0. Now plug those values into the equation 1%2Ax-3%2Ay=0

1%2A%280%29-3%2A%28%2B0%29=0 Plug in x=0 and y=%2B0


0%2B0=0 Multiply


0=0 Add


0=0 Reduce. Since this equation is true the solution works.


So the solution (0,%2B0) satisfies 1%2Ax-3%2Ay=0


Since the solution (0,%2B0) satisfies the system of equations


5%2Ax%2B2%2Ay=0
1%2Ax-3%2Ay=0


this verifies our answer.




Circles/113898: Give the equation for the circle with center C(3,-2) and radius 4.
1 solutions

Answer 82881 by jim_thompson5910(28598) About Me  on 2007-12-02 12:05:10 (Show Source):
You can put this solution on YOUR website!
The general equation of a circle is %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 where (h,k) is the center and r is the radius.

So using this equation, we get

%28x-3%29%5E2%2B%28y-%28-2%29%29%5E2=4%5E2 Plug in h=3,k=-2, and r=4. These are given


%28x-3%29%5E2%2B%28y%2B2%29%5E2=4%5E2 Rewrite y-%28-2%29 as y%2B2

%28x-3%29%5E2%2B%28y%2B2%29%5E2=16 Square 4 to get 16


So the equation for the circle with center C(3,-2) and radius 4 is %28x-3%29%5E2%2B%28y%2B2%29%5E2=16


Coordinate-system/113889: Write the equation of the line that passes through point (–7, –15) with a slope of 3.

1 solutions

Answer 82880 by jim_thompson5910(28598) About Me  on 2007-12-02 12:01:35 (Show Source):
You can put this solution on YOUR website!

If you want to find the equation of line with a given a slope of 3 which goes through the point (-7,-15), you can simply use the point-slope formula to find the equation:


---Point-Slope Formula---
y-y%5B1%5D=m%28x-x%5B1%5D%29 where m is the slope, and is the given point

So lets use the Point-Slope Formula to find the equation of the line

y--15=%283%29%28x--7%29 Plug in m=3, x%5B1%5D=-7, and y%5B1%5D=-15 (these values are given)


y%2B15=%283%29%28x--7%29 Rewrite y--15 as y%2B15


y%2B15=%283%29%28x%2B7%29 Rewrite x--7 as x%2B7


y%2B15=3x%2B%283%29%287%29 Distribute 3

y%2B15=3x%2B21 Multiply 3 and 7 to get 21

y=3x%2B21-15 Subtract 15 from both sides to isolate y

y=3x%2B6 Combine like terms 21 and -15 to get 6
------------------------------------------------------------------------------------------------------------
Answer:


So the equation of the line with a slope of 3 which goes through the point (-7,-15) is:

y=3x%2B6 which is now in y=mx%2Bb form where the slope is m=3 and the y-intercept is b=6

Notice if we graph the equation y=3x%2B6 and plot the point (-7,-15), we get (note: if you need help with graphing, check out this solver)

Graph of y=3x%2B6 through the point (-7,-15)
and we can see that the point lies on the line. Since we know the equation has a slope of 3 and goes through the point (-7,-15), this verifies our answer.


Coordinate-system/113888: Write the equation of the line with slope 4 and y-intercept (0, –5). Then graph the line.
1 solutions

Answer 82879 by jim_thompson5910(28598) About Me  on 2007-12-02 12:00:34 (Show Source):
You can put this solution on YOUR website!
If we have the y-intercept (0, –5), then the line goes through the point (0, –5)




If you want to find the equation of line with a given a slope of 4 which goes through the point (0,-5), you can simply use the point-slope formula to find the equation:


---Point-Slope Formula---
y-y%5B1%5D=m%28x-x%5B1%5D%29 where m is the slope, and is the given point

So lets use the Point-Slope Formula to find the equation of the line

y--5=%284%29%28x-0%29 Plug in m=4, x%5B1%5D=0, and y%5B1%5D=-5 (these values are given)


y%2B5=%284%29%28x-0%29 Rewrite y--5 as y%2B5


y%2B5=4x%2B%284%29%28-0%29 Distribute 4

y%2B5=4x%2B0 Multiply 4 and -0 to get 0

y=4x%2B0-5 Subtract 5 from both sides to isolate y

y=4x-5 Combine like terms 0 and -5 to get -5
------------------------------------------------------------------------------------------------------------
Answer:


So the equation of the line with a slope of 4 which goes through the point (0,-5) is:

y=4x-5 which is now in y=mx%2Bb form where the slope is m=4 and the y-intercept is b=-5

Notice if we graph the equation y=4x-5 and plot the point (0,-5), we get (note: if you need help with graphing, check out this solver)

Graph of y=4x-5 through the point (0,-5)
and we can see that the point lies on the line. Since we know the equation has a slope of 4 and goes through the point (0,-5), this verifies our answer.


Coordinate-system/113890: Write the equation of the line passing through (6, –5) and (–3, 4)
1 solutions

Answer 82878 by jim_thompson5910(28598) About Me  on 2007-12-02 11:59:08 (Show Source):
You can put this solution on YOUR website!
First lets find the slope through the points (6,-5) and (-3,4)

m=%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29 Start with the slope formula (note: is the first point (6,-5) and is the second point (-3,4))

m=%284--5%29%2F%28-3-6%29 Plug in y%5B2%5D=4,y%5B1%5D=-5,x%5B2%5D=-3,x%5B1%5D=6 (these are the coordinates of given points)

m=+9%2F-9 Subtract the terms in the numerator 4--5 to get 9. Subtract the terms in the denominator -3-6 to get -9


m=-1 Reduce

So the slope is
m=-1

------------------------------------------------


Now let's use the point-slope formula to find the equation of the line:



------Point-Slope Formula------
y-y%5B1%5D=m%28x-x%5B1%5D%29 where m is the slope, and is one of the given points

So lets use the Point-Slope Formula to find the equation of the line

y--5=%28-1%29%28x-6%29 Plug in m=-1, x%5B1%5D=6, and y%5B1%5D=-5 (these values are given)


y%2B5=%28-1%29%28x-6%29 Rewrite y--5 as y%2B5


y%2B5=-x%2B%28-1%29%28-6%29 Distribute -1

y%2B5=-x%2B6 Multiply -1 and -6 to get 6

y=-x%2B6-5 Subtract 5 from both sides to isolate y

y=-x%2B1 Combine like terms 6 and -5 to get 1
------------------------------------------------------------------------------------------------------------
Answer:


So the equation of the line which goes through the points (6,-5) and (-3,4) is:y=-x%2B1

The equation is now in y=mx%2Bb form (which is slope-intercept form) where the slope is m=-1 and the y-intercept is b=1

Notice if we graph the equation y=-x%2B1 and plot the points (6,-5) and (-3,4), we get this: (note: if you need help with graphing, check out this solver)

Graph of y=-x%2B1 through the points (6,-5) and (-3,4)

Notice how the two points lie on the line. This graphically verifies our answer.


Complex_Numbers/113906: Can I take average of a complex number and a real number?
i.e. [(a+bi)+c]/2=?
1 solutions

Answer 82877 by jim_thompson5910(28598) About Me  on 2007-12-02 11:57:06 (Show Source):
You can put this solution on YOUR website!
If you had something like %28%28a%2Bbi%29%2Bc%29%2F2 then you will get a complex number as a result. So you'll get %28x%2Bbi%29%2F2 where x=a%2Bc since you can only add the real parts.


Human-and-algebraic-language/113900: 1. solve these equations
y=10-x
2x-y=-4
1 solutions

Answer 82876 by jim_thompson5910(28598) About Me  on 2007-12-02 11:54:49 (Show Source):
You can put this solution on YOUR website!
Start with the given system
2x-y=-4
y=10-x



2x-%2810-x%29=-4 Plug in y=10-x into the first equation. In other words, replace each y with 10-x. Notice we've eliminated the y variables. So we now have a simple equation with one unknown.


2x-10%2Bx=-4 Distribute the negative


3x-10=-4 Combine like terms on the left side


3x=-4%2B10Add 10 to both sides


3x=6 Combine like terms on the right side


x=%286%29%2F%283%29 Divide both sides by 3 to isolate x



x=2 Divide




Now that we know that x=2, we can plug this into y=10-x to find y



y=10-%282%29 Substitute 2 for each x


y=8 Simplify


So our answer is x=2 and y=8


Linear-equations/113894: x+y=4
-x+y=2
solve using addition
1 solutions

Answer 82875 by jim_thompson5910(28598) About Me  on 2007-12-02 11:49:52 (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations

1%2Ax%2B1%2Ay=4
-1%2Ax%2B1%2Ay=2

In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa).

So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero.

So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get 1 and -1 to some equal number, we could try to get them to the LCM.

Since the LCM of 1 and -1 is -1, we need to multiply both sides of the top equation by -1 and multiply both sides of the bottom equation by -1 like this:

-1%2A%281%2Ax%2B1%2Ay%29=%284%29%2A-1 Multiply the top equation (both sides) by -1
-1%2A%28-1%2Ax%2B1%2Ay%29=%282%29%2A-1 Multiply the bottom equation (both sides) by -1


So after multiplying we get this:
-1%2Ax-1%2Ay=-4
1%2Ax-1%2Ay=-2

Notice how -1 and 1 add to zero (ie -1%2B1=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%28-1%2Ax%2B1%2Ax%29-1%2Ay-1%2Ay%29=-4-2

%28-1%2B1%29%2Ax-1-1%29y=-4-2

cross%28-1%2B1%29%2Ax%2B%28-1-1%29%2Ay=-4-2 Notice the x coefficients add to zero and cancel out. This means we've eliminated x altogether.



So after adding and canceling out the x terms we're left with:

-2%2Ay=-6

y=-6%2F-2 Divide both sides by -2 to solve for y



y=3 Reduce


Now plug this answer into the top equation 1%2Ax%2B1%2Ay=4 to solve for x

1%2Ax%2B1%283%29=4 Plug in y=3


1%2Ax%2B3=4 Multiply



1%2Ax=4-3 Subtract 3 from both sides

1%2Ax=1 Combine the terms on the right side

cross%28%281%2F1%29%281%29%29%2Ax=%281%29%281%2F1%29 Multiply both sides by 1%2F1. This will cancel out 1 on the left side.


x=1 Multiply the terms on the right side


So our answer is

x=1, y=3

which also looks like

(1, 3)

Notice if we graph the equations (if you need help with graphing, check out this solver)

1%2Ax%2B1%2Ay=4
-1%2Ax%2B1%2Ay=2

we get



graph of 1%2Ax%2B1%2Ay=4 (red) -1%2Ax%2B1%2Ay=2 (green) (hint: you may have to solve for y to graph these) and the intersection of the lines (blue circle).


and we can see that the two equations intersect at (1,3). This verifies our answer.


Human-and-algebraic-language/113901: 5. solve the following systems of equations
3x+5y=25
x+2y=10
1 solutions

Answer 82874 by jim_thompson5910(28598) About Me  on 2007-12-02 11:48:29 (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Solving a linear system of equations by subsitution


Lets start with the given system of linear equations

3%2Ax%2B5%2Ay=25
1%2Ax%2B2%2Ay=10

Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to choose y.

Solve for y for the first equation

5%2Ay=25-3%2AxSubtract 3%2Ax from both sides

y=%2825-3%2Ax%29%2F5 Divide both sides by 5.


Which breaks down and reduces to



y=5-%283%2F5%29%2Ax Now we've fully isolated y

Since y equals 5-%283%2F5%29%2Ax we can substitute the expression 5-%283%2F5%29%2Ax into y of the 2nd equation. This will eliminate y so we can solve for x.


1%2Ax%2B2%2Ahighlight%28%285-%283%2F5%29%2Ax%29%29=10 Replace y with 5-%283%2F5%29%2Ax. Since this eliminates y, we can now solve for x.

1%2Ax%2B2%2A%285%29%2B2%28-3%2F5%29x=10 Distribute 2 to 5-%283%2F5%29%2Ax

1%2Ax%2B10-%286%2F5%29%2Ax=10 Multiply



1%2Ax%2B10-%286%2F5%29%2Ax=10 Reduce any fractions

1%2Ax-%286%2F5%29%2Ax=10-10 Subtract 10 from both sides


1%2Ax-%286%2F5%29%2Ax=0 Combine the terms on the right side



%285%2F5%29%2Ax-%286%2F5%29x=0 Make 1 into a fraction with a denominator of 5

%28-1%2F5%29%2Ax=0 Now combine the terms on the left side.


cross%28%285%2F-1%29%28-1%2F5%29%29x=%280%2F1%29%285%2F-1%29 Multiply both sides by 5%2F-1. This will cancel out -1%2F5 and isolate x

So when we multiply 0%2F1 and 5%2F-1 (and simplify) we get



x=0 <---------------------------------One answer

Now that we know that x=0, lets substitute that in for x to solve for y

1%280%29%2B2%2Ay=10 Plug in x=0 into the 2nd equation

0%2B2%2Ay=10 Multiply

2%2Ay=10%2B0Add 0 to both sides

2%2Ay=10 Combine the terms on the right side

cross%28%281%2F2%29%282%29%29%2Ay=%2810%2F1%29%281%2F2%29 Multiply both sides by 1%2F2. This will cancel out 2 on the left side.

y=10%2F2 Multiply the terms on the right side


y=5 Reduce


So this is the other answer


y=5<---------------------------------Other answer


So our solution is

x=0 and y=5

which can also look like

(0,5)

Notice if we graph the equations (if you need help with graphing, check out this solver)

3%2Ax%2B5%2Ay=25
1%2Ax%2B2%2Ay=10

we get


graph of 3%2Ax%2B5%2Ay=25 (red) and 1%2Ax%2B2%2Ay=10 (green) (hint: you may have to solve for y to graph these) intersecting at the blue circle.


and we can see that the two equations intersect at (0,5). This verifies our answer.


-----------------------------------------------------------------------------------------------
Check:

Plug in (0,5) into the system of equations


Let x=0 and y=5. Now plug those values into the equation 3%2Ax%2B5%2Ay=25

3%2A%280%29%2B5%2A%285%29=25 Plug in x=0 and y=5


0%2B25=25 Multiply


25=25 Add


25=25 Reduce. Since this equation is true the solution works.


So the solution (0,5) satisfies 3%2Ax%2B5%2Ay=25



Let x=0 and y=5. Now plug those values into the equation 1%2Ax%2B2%2Ay=10

1%2A%280%29%2B2%2A%285%29=10 Plug in x=0 and y=5


0%2B10=10 Multiply


10=10 Add


10=10 Reduce. Since this equation is true the solution works.


So the solution (0,5) satisfies 1%2Ax%2B2%2Ay=10


Since the solution (0,5) satisfies the system of equations


3%2Ax%2B5%2Ay=25
1%2Ax%2B2%2Ay=10


this verifies our answer.




Human-and-algebraic-language/113902: 5. solve the following systems of equations
x=3y+7
x+y=-5
1 solutions

Answer 82873 by jim_thompson5910(28598) About Me  on 2007-12-02 11:47:09 (Show Source):
You can put this solution on YOUR website!
Start with the given system
x%2By=-5
x=3y%2B7



%283y%2B7%29%2By=-5 Plug in x=3y%2B7 into the first equation. In other words, replace each x with 3y%2B7. Notice we've eliminated the x variables. So we now have a simple equation with one unknown.


3y%2B7%2By=-5


4y%2B7=-5 Combine like terms on the left side


4y=-5-7Subtract 7 from both sides


4y=-12 Combine like terms on the right side


y=%28-12%29%2F%284%29 Divide both sides by 4 to isolate y



y=-3 Divide




Now that we know that y=-3, we can plug this into x=3y%2B7 to find x



x=3%28-3%29%2B7 Substitute -3 for each y


x=-2 Simplify


So our answer is x=-2 and y=-3


Human-and-algebraic-language/113903: 8. solve the following systems of equations
y=3x-5
6x-3y=3
1 solutions

Answer 82872 by jim_thompson5910(28598) About Me  on 2007-12-02 11:45:18 (Show Source):
You can put this solution on YOUR website!

Start with the given system
6x-3y=3
y=3x-5



6x-3%283x-5%29=3 Plug in y=3x-5 into the first equation. In other words, replace each y with 3x-5. Notice we've eliminated the y variables. So we now have a simple equation with one unknown.


6x-9x%2B15=3 Distribute


-3x%2B15=3 Combine like terms on the left side


-3x=3-15Subtract 15 from both sides


-3x=-12 Combine like terms on the right side


x=%28-12%29%2F%28-3%29 Divide both sides by -3 to isolate x



x=4 Divide




Now that we know that x=4, we can plug this into y=3x-5 to find y



y=3%284%29-5 Substitute 4 for each x


y=7 Simplify


So our answer is x=4 and y=7


Linear-equations/113838: I am having trouble with linear equations by elimination.
x + y = 9
x + 4y = 15
I need to find x and y
Thank you very much
1 solutions

Answer 82858 by jim_thompson5910(28598) About Me  on 2007-12-02 00:16:12 (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations

1%2Ax%2B1%2Ay=9
1%2Ax%2B4%2Ay=15

In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa).

So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero.

So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get 1 and 1 to some equal number, we could try to get them to the LCM.

Since the LCM of 1 and 1 is 1, we need to multiply both sides of the top equation by 1 and multiply both sides of the bottom equation by -1 like this:

1%2A%281%2Ax%2B1%2Ay%29=%289%29%2A1 Multiply the top equation (both sides) by 1
-1%2A%281%2Ax%2B4%2Ay%29=%2815%29%2A-1 Multiply the bottom equation (both sides) by -1


So after multiplying we get this:
1%2Ax%2B1%2Ay=9
-1%2Ax-4%2Ay=-15

Notice how 1 and -1 add to zero (ie 1%2B-1=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%281%2Ax-1%2Ax%29%2B%281%2Ay-4%2Ay%29=9-15

%281-1%29%2Ax%2B%281-4%29y=9-15

cross%281%2B-1%29%2Ax%2B%281-4%29%2Ay=9-15 Notice the x coefficients add to zero and cancel out. This means we've eliminated x altogether.



So after adding and canceling out the x terms we're left with:

-3%2Ay=-6

y=-6%2F-3 Divide both sides by -3 to solve for y



y=2 Reduce


Now plug this answer into the top equation 1%2Ax%2B1%2Ay=9 to solve for x

1%2Ax%2B1%282%29=9 Plug in y=2


1%2Ax%2B2=9 Multiply



1%2Ax=9-2 Subtract 2 from both sides

1%2Ax=7 Combine the terms on the right side

cross%28%281%2F1%29%281%29%29%2Ax=%287%29%281%2F1%29 Multiply both sides by 1%2F1. This will cancel out 1 on the left side.


x=7 Multiply the terms on the right side


So our answer is

x=7, y=2

which also looks like

(7, 2)

Notice if we graph the equations (if you need help with graphing, check out this solver)

1%2Ax%2B1%2Ay=9
1%2Ax%2B4%2Ay=15

we get



graph of 1%2Ax%2B1%2Ay=9 (red) 1%2Ax%2B4%2Ay=15 (green) (hint: you may have to solve for y to graph these) and the intersection of the lines (blue circle).


and we can see that the two equations intersect at (7,2). This verifies our answer.


Coordinate-system/113842: Determine which two equations represent parallel lines.
(a) y = 2x – 6 (b) y = –2x + 6 (c) y = 2x + 1 (d) y = -1/2 x – 6



1 solutions

Answer 82855 by jim_thompson5910(28598) About Me  on 2007-12-01 22:30:49 (Show Source):
You can put this solution on YOUR website!
Since A) and C) have equal slopes of 2 (the slope is the number in front of the x), these two lines are parallel.


Coordinate-system/113839: graph 3x + y = 3
1 solutions

Answer 82853 by jim_thompson5910(28598) About Me  on 2007-12-01 22:29:45 (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Graphing Linear Equations


3%2Ax%2B1%2Ay=3Start with the given equation



1%2Ay=3-3%2Ax Subtract 3%2Ax from both sides

y=%281%29%283-3%2Ax%29 Multiply both sides by 1

y=%281%29%283%29-%281%29%283%29x%29 Distribute 1

y=3-%283%29x Multiply

y=-3%2Ax%2B3 Rearrange the terms

y=-3%2Ax%2B3 Reduce any fractions

So the equation is now in slope-intercept form (y=mx%2Bb) where m=-3 (the slope) and b=3 (the y-intercept)

So to graph this equation lets plug in some points

Plug in x=-2

y=-3%2A%28-2%29%2B3

y=6%2B3 Multiply

y=9 Add

So here's one point (-2,9)





Now lets find another point

Plug in x=-1

y=-3%2A%28-1%29%2B3

y=3%2B3 Multiply

y=6 Add

So here's another point (-1,6). Add this to our graph





Now draw a line through these points

So this is the graph of y=-3%2Ax%2B3 through the points (-2,9) and (-1,6)


So from the graph we can see that the slope is -3%2F1 (which tells us that in order to go from point to point we have to start at one point and go down -3 units and to the right 1 units to get to the next point), the y-intercept is (0,3)and the x-intercept is (1,0) . So all of this information verifies our graph.


We could graph this equation another way. Since b=3 this tells us that the y-intercept (the point where the graph intersects with the y-axis) is (0,3).


So we have one point (0,3)






Now since the slope is -3%2F1, this means that in order to go from point to point we can use the slope to do so. So starting at (0,3), we can go down 3 units


and to the right 1 units to get to our next point



Now draw a line through those points to graph y=-3%2Ax%2B3


So this is the graph of y=-3%2Ax%2B3 through the points (0,3) and (1,0)


Coordinate-system/113841: Find the slope of the line passing through the points (5, 3) and (6, 7)
1 solutions

Answer 82852 by jim_thompson5910(28598) About Me  on 2007-12-01 22:28:55 (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Finding the slope


Slope of the line through the points (5, 3) and (6, 7)



m+=+%28y%5B2%5D+-+y%5B1%5D%29%2F%28x%5B2%5D+-+x%5B1%5D%29


m+=+%287+-+3%29%2F%286+-+5%29


m+=+%284%29%2F%281%29


m+=+4



Answer: Slope is m+=+4



Coordinate-system/113840: Graph using the intercept method: 2x – 5y = 10
1 solutions

Answer 82851 by jim_thompson5910(28598) About Me  on 2007-12-01 22:28:16 (Show Source):
You can put this solution on YOUR website!

2%2Ax-5%2Ay=10 Start with the given equation

Let's find the x-intercept

To find the x-intercept, let y=0 and solve for x:
2%2Ax-5%2A%280%29=10 Plug in y=0

2%2Ax=10 Simplify

x=10%2F2 Divide both sides by 2


x=5 Reduce



So the x-intercept is (note: the x-intercept will always have a y-coordinate equal to zero)



------------------

2%2Ax-5%2Ay=10 Start with the given equation

Now let's find the y-intercept

To find the y-intercept, let x=0 and solve for y:
2%2A%280%29-5%2Ay=10 Plug in x=0

5%2Ay=10 Simplify

x=10%2F-5 Divide both sides by -5



y=-2 Reduce



So the y-intercept is (note: the y-intercept will always have a x-coordinate equal to zero)

------------------------------------------

So we have these intercepts:
x-intercept:

y-intercept:



Now plot the two points and




Now draw a line through the two points to graph 2%2Ax-5%2Ay=10
graph of 2%2Ax-5%2Ay=10 through the points and


Polynomials-and-rational-expressions/113829: 10. Solve the following system both by graphing and then check result algebraically.
y = 8
2y = x^2

1 solutions

Answer 82837 by jim_thompson5910(28598) About Me  on 2007-12-01 21:45:24 (Show Source):
You can put this solution on YOUR website!
First let's graph y=8

In order to graph y=8, simply draw a horizontal line through the y=8

+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+8%29+ Graph of y=8


Now let's move onto the second equation 2y=x%5E2

2y=x%5E2 Start with the given equation


y=x%5E2%2F2 Divide both sides by 2 to solve for y


Now let's graph y=x%5E2%2F2


In order to do so, let's make a table by plugging in x-values (you get to choose which ones) to find the y-values


       x          y
   -5.00000   12.50000
   -4.00000    8.00000
   -3.00000    4.50000
   -2.00000    2.00000
   -1.00000    0.50000
    0.00000    0.00000
    1.00000    0.50000
    2.00000    2.00000
    3.00000    4.50000
    4.00000    8.00000
    5.00000   12.50000




Now let's plot these points and connect them to get this graph


+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E2%2F2%29+ Graph of y=x%5E2%2F2


Now let's plot the two graphs together:

+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C8%2C+x%5E2%2F2%29+ Graph of y=8 (red) y=x%5E2%2F2 (green)


From the graph, we can see that the two lines intersect at x=-4 and x=4 which means the solutions are (-4,8) or (4,8)





=====================================


Now let's solve algebraically:

Start with the given system
y=8
2y=x%5E2


2%288%29=x%5E2 Take the 2nd equation and plug in y=8


16=x%5E2 Multiply


Take the square root of both sides


So we then get x=-4 or x=4

So this means our solutions are (-4,8) or (4,8) (remember y is always 8 since the first equation is y=8)




Polynomials-and-rational-expressions/113827: 9. Evaluate the following:
a. log1/2 4
b. 1og8 64
c. log32 8
d. 1og3 81
e. pi^(log_pi 2pi)
1 solutions

Answer 82836 by jim_thompson5910(28598) About Me  on 2007-12-01 21:43:52 (Show Source):
You can put this solution on YOUR website!
Remember if we have log%28b%2C%28x%29%29=y, we can write it as b%5Ey=x


a)
log%281%2F2%2C%284%29%29=y Start with the given equation. Just set the original expression equal to y. Note: the entire 1%2F2 is the base of the log.


%281%2F2%29%5Ey=4 Rewrite the original expression using the property I listed above



%282%5E%28-1%29%29%5Ey=4 Rewrite 1%2F2 as 2%5E%28-1%29


2%5E%28-y%29=4 Multiply the exponents



2%5E%28-y%29=2%5E2 Rewrite 4 as 2%5E2


Since the bases are equal, the exponents are equal. So -y=2 which means y=-2

So log%281%2F2%2C%284%29%29=-2


-------------------------------------


b)
log%288%2C%2864%29%29=y Start with the given equation. Just set the original expression equal to y


8%5Ey=64 Rewrite the original expression using the property: log%28b%2C%28x%29%29=y <===> b%5Ey=x


8%5Ey=8%5E2 Rewrite 64 as 8%5E2


Since the bases are equal, the exponents are equal. So y=2

So log%288%2C%2864%29%29=2


---------------------------------------
c)

log%2832%2C%288%29%29=y Start with the given equation. Just set the original expression equal to y


32%5Ey=8 Rewrite the original expression using the property: log%28b%2C%28x%29%29=y <===> b%5Ey=x


%282%5E5%29%5Ey=2%5E3 Rewrite 8 as 2%5E3 and 32 as 2%5E5


2%5E%285y%29=2%5E3 Multiply the exponents

Since the bases are equal, the exponents are equal. So 5y=3 which means y=3%2F5

So log%2832%2C%288%29%29=3%2F5

---------------------------------------
d)

log%283%2C%2881%29%29=y Start with the given equation. Just set the original expression equal to y


3%5Ey=81 Rewrite the original expression using the property: log%28b%2C%28x%29%29=y <===> b%5Ey=x


3%5Ey=3%5E4 Rewrite 81 as 3%5E4

Since the bases are equal, the exponents are equal. So y=4

So log%283%2C%2881%29%29=4

---------------------------------------
e)

pi%5E%28log%28pi%2C%282pi%29%29%29 Start with the given expression


pi%5E%28log%28pi%2C%282%29%29%2Blog%28pi%2C%28pi%29%29%29 Break up the logarithm using the identity log%28b%2C%28x%2Ay%29%29=log%28b%2C%28x%29%29%2Blog%28b%2C%28y%29%29. Think of 2pi as 2 times pi.


pi%5E%28log%28pi%2C%282%29%29%29%2Api%5E%28log%28pi%2C%28pi%29%29%29 Break up the exponent using the identity b%5E%28x%2By%29=b%5Ex%2Ab%5Ey


pi%5E%28log%28pi%2C%282%29%29%29%2Api%5E1 Evaluate log%28pi%2C%28pi%29%29 to get 1

Since we cannot simplify the expression log%28pi%2C%282%29%29 any further, we cannot simplify the entire expression any further.




Polynomials-and-rational-expressions/113826: 8. Solve and Check the Following Equations:
C. log6 x + log6 (x – 2) = log6 15
1 solutions

Answer 82835 by jim_thompson5910(28598) About Me  on 2007-12-01 21:41:30 (Show Source):
You can put this solution on YOUR website!
log%286%2C%28x%29%29%2Blog%286%2C%28x-2%29%29=log%286%2C%2815%29%29 Start with the given equation


log%286%2C%28x%28x-2%29%29%29=log%286%2C%2815%29%29 Combine the logs using the identity log%28b%2C%28x%29%29%2Blog%28b%2C%28y%29%29=log%28b%2C%28x%2Ay%29%29



log%286%2C%28x%5E2-2x%29%29=log%286%2C%2815%29%29 Distribute


Since the base of the logs are equal, the arguments (the stuff inside the logs are equal). So x%5E2-2x=15


So let's solve x%5E2-2x=15

x%5E2-2x=15 Start with the given equation

x%5E2-2x-15=0 Subtract 15 from both sides


%28x-5%29%28x%2B3%29=0 Factor the left side (note: if you need help with factoring, check out this
solver)



Now set each factor equal to zero:

x-5=0 or x%2B3=0

x=5 or x=-3 Now solve for x in each case


So our possible solutions are x=5 or x=-3


However, since you cannot take the log of a negative number, the only solution is x=5

-------------------------------------
Check:

Let's check the solution x=5

log%286%2C%28x%5E2-2x%29%29=log%286%2C%2815%29%29 Start with the equation found on the third step.


log%286%2C%285%5E2-2%285%29%29%29=log%286%2C%2815%29%29 Plug in x=5



log%286%2C%2825-10%29%29=log%286%2C%2815%29%29 Square and multiply


log%286%2C%2815%29%29=log%286%2C%2815%29%29 Subtract. Since both sides of the equation are equal, this solution is verified.



So the solution x=5 is verified


Polynomials-and-rational-expressions/113823: 8. Solve and Check the Following Equations:
A. 2^(x^2+x) = 64

1 solutions

Answer 82833 by jim_thompson5910(28598) About Me  on 2007-12-01 21:35:55 (Show Source):
You can put this solution on YOUR website!
2%5E%28x%5E2%2Bx%29=64 Start with the given equation


2%5E%28x%5E2%2Bx%29=2%5E6 Rewrite 64 as 2%5E6


Since the bases are equal, the exponents are equal. So x%5E2%2Bx=6

x%5E2%2Bx=6 Now let's solve for x


x%5E2%2Bx-6=0 Subtract x from both sides



%28x%2B3%29%28x-2%29=0 Factor the left side (note: if you need help with factoring, check out this
solver)



Now set each factor equal to zero:

x%2B3=0 or x-2=0

x=-3 or x=2 Now solve for x in each case


So our solutions are x=-3 or x=2


------------------------

Check:
Let's check the solution x=-3

2%5E%28x%5E2%2Bx%29=64 Start with the given equation


2%5E%28%28-3%29%5E2%2B%28-3%29%29=64 Plug in x=-3


2%5E%289%2B%28-3%29%29=64 Square -3 to get 9


2%5E6=64 Add

64=64 Raise 2 to the sixth power to get 64. Since both sides of the equation are equal, this solution is verified.

----




Let's check the solution x=2

2%5E%28x%5E2%2Bx%29=64 Start with the given equation


2%5E%28%282%29%5E2%2B%282%29%29=64 Plug in x=2


2%5E%289%2B%282%29%29=64 Square 2 to get 4


2%5E6=64 Add

64=64 Raise 2 to the sixth power to get 64. Since both sides of the equation are equal, this solution is verified.


So this verifies our solutions x=-3 and x=2


Polynomials-and-rational-expressions/113822: 7. Graph f%28x%29+=+3%5Ex and it's inverse on the same set of coordinate axes. Label all units, intercepts, and asymptotes. Determine the domain and range of each function.
1 solutions

Answer 82831 by jim_thompson5910(28598) About Me  on 2007-12-01 21:32:19 (Show Source):
You can put this solution on YOUR website!
+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+3%5Ex%2C+log%283%2Cx%29%29+ Graph of 3%5Ex (red) and it's inverse log%283%2C%28x%29%29 (green)


So we can see from the graph of 3%5Ex we can see...

f:

Domain: all real numbers,

Range=,

Asymptotes: no vertical, horizontal at y=0

Intercepts: x-int=none, y-int=(0,1)



And we can see from the graph of log%283%2C%28x%29%29 we can see...

f%5E%28-1%29:

Domain: ,

Range=all real numbers,

Asymptotes: vertical at x=0, no horizontal,

Intercepts: x-int=(1,0) y-int=none


Notice how everything switched from the graphs f to f%5E%28-1%29


Polynomials-and-rational-expressions/113821: 6. Determine the behavior of f%28x%29=%28x%5E2%2Bx-6%29%2F%28x-3%29

Domain :
x – intercept(s) :
Hole(s) :
y – intercept :
Vertical Asymptote(s) :
Oblique Asymptote :

1 solutions

Answer 82830 by jim_thompson5910(28598) About Me  on 2007-12-01 21:29:13 (Show Source):
You can put this solution on YOUR website!
Domain:


f%28x%29=%28x%5E2%2Bx-6%29%2F%28x-3%29 Start with the given function


x-3=0 Set the denominator equal to zero. Remember, dividing by 0 is undefined. So if we find values of x that make the
denominator zero, then we must exclude them from the domain.



x=0%2B3Add 3 to both sides


x=3 Combine like terms on the right side





Since x=3 makes the denominator equal to zero, this means we must exclude x=3 from our domain


Answer:
So our domain is:

which in plain English reads: x is the set of all real numbers except x%3C%3E3

So our domain looks like this in interval notation






x-intercept(s):

f%28x%29=%28x%5E2%2Bx-6%29%2F%28x-3%29 Start with the given function


0=%28x%5E2%2Bx-6%29%2F%28x-3%29 To find the x-intercepts, set f(x) (which is y) equal to zero


x%5E2%2Bx-6=0 This means the numerator is equal to zero. Remember the denominator cannot equal zero



%28x%2B3%29%28x-2%29=0 Factor the left side (note: if you need help with factoring, check out this
solver)



Now set each factor equal to zero:

x%2B3=0 or x-2=0

x=-3 or x=2 Now solve for x in each case


Answer:
So our solutions are x=-3 or x=2
This means the x-intercepts are (-3,0) and (2,0)




Hole(s):

f%28x%29=%28x%5E2%2Bx-6%29%2F%28x-3%29 Start with the given function


f%28x%29=%28x%2B3%29%28x-2%29%2F%28x-3%29 Factor


Answer:
Since the expression does not simplify further, there are no exceptions to make about the two expressions. So in this case there are no holes in this function





y-intercept(s):


f%28x%29=%28x%5E2%2Bx-6%29%2F%28x-3%29 Start with the given function


f%280%29=%280%5E2%2B0-6%29%2F%280-3%29 To find the y-intercept, plug in x=0


f%280%29=%28-6%29%2F%28-3%29 Simplify


f%280%29=2 Divide


Answer:

So the y intercept is (0,2)







Vertical Asymptote(s):


Since the value x=3 is excluded from the domain, and there are no holes, there is one vertical asymptote at x=3

Answer:

So the vertical asymptote is x=3





Oblique Asymptote:

To find the oblique asymptote, divide %28x%5E2%2Bx-6%29%2F%28x-3%29 using synthetic division


Start with the given expression %28x%5E2+%2B+x+-+6%29%2F%28x-3%29

First lets find our test zero:

x-3=0 Set the denominator x-3 equal to zero

x=3 Solve for x.

so our test zero is 3


Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.
3|11-6
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
3|11-6
|
1

Multiply 3 by 1 and place the product (which is 3) right underneath the second coefficient (which is 1)
3|11-6
|3
1

Add 3 and 1 to get 4. Place the sum right underneath 3.
3|11-6
|3
14

Multiply 3 by 4 and place the product (which is 12) right underneath the third coefficient (which is -6)
3|11-6
|312
14

Add 12 and -6 to get 6. Place the sum right underneath 12.
3|11-6
|312
146

Since the last column adds to 6, we have a remainder of 6. This means x-3 is not a factor of x%5E2+%2B+x+-+6
Now lets look at the bottom row of coefficients:

The first 2 coefficients (1,4) form the quotient

x+%2B+4




So the oblique asymptote is the line y=x%2B4



Notice if we graph f%28x%29=%28x%5E2%2Bx-6%29%2F%28x-3%29 and it's asymptotes, we can visually verify our answers:


Graph of f%28x%29=%28x%5E2%2Bx-6%29%2F%28x-3%29 with the oblique asymptote y=x%2B4 (green) and the vertical asymptote x=3 (blue)