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Start with the given equation



Take the square root of both sides




Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



Subtract 7 from both sides to isolate x.


Break down the expression into two parts:

   or  

 

Now combine like terms for each expression:

   or      

-----------------------------------
Answer:
So our solution is

  or   

 

Notice when we graph the equations and we get:

graph of (red) and (green)



Here we can see that the two equations intersect at x values of and , so this verifies our answer.

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a)

Let's find f(2)

Start with the given function


Plug in


Raise 2 to the 2nd power to get 4


Multiply 2 and 4 to get 8


Multiply 7 and 2 to get 14


Subtract 14 from 8 to get -6


Add -6 and 3 to get -3


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

b)

Let's find f(-3)

Start with the given function


Plug in


Raise -3 to the 2nd power to get 9


Multiply 2 and 9 to get 18


Multiply 7 and -3 to get -21


Subtract -21 from 18 to get 39


Add 39 and 3 to get 42




So you are correct.

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From the quadratic formula


the discriminant consists of all of the terms in the square root. So the discriminant is



the discriminant tells us how many solutions (and what type of solutions) we can expect for any quadratic.


Now let's find the discriminant for (notice how , and ):

Start with the given equation

Plug in a=1, b=1, c=3

Square 1 to get 1

Multiply -4*1*3 to get -12

Combine 1 and -12 to get -11


Since the discriminant equals -11 (which is less than zero) , this means there are two complex solutions.


Notice if we graph , we get



and we can see that there are two complex solutions. So this visually verifies our answer.

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Start with the given equation

To find the y-intercept, let x=0 and solve for y:
Plug in

Simplify

Divide both sides by 3 to isolate y


Reduce



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

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In order to generate ordered pairs, we need to plot evaluate some x values to get some y values to get the points.


So let's find the first point:



Start with the given function


Plug in


Raise -5 to the 2nd power to get 25


Multiply 1 and 25 to get 25


Subtract 3 from 25 to get 22


So when , we have


So our 1st point is (-5,22)


-------------- Let's find another point --------------


Start with the given function


Plug in


Raise -4 to the 2nd power to get 16


Multiply 1 and 16 to get 16


Subtract 3 from 16 to get 13


So when , we have


So our 2nd point is (-4,13)


-------------- Let's find another point --------------


Start with the given function


Plug in


Raise -3 to the 2nd power to get 9


Multiply 1 and 9 to get 9


Subtract 3 from 9 to get 6


So when , we have


So our 3rd point is (-3,6)


-------------- Let's find another point --------------


Start with the given function


Plug in


Raise -2 to the 2nd power to get 4


Multiply 1 and 4 to get 4


Subtract 3 from 4 to get 1


So when , we have


So our 4th point is (-2,1)


-------------- Let's find another point --------------


Start with the given function


Plug in


Raise -1 to the 2nd power to get 1


Multiply 1 and 1 to get 1


Subtract 3 from 1 to get -2


So when , we have


So our 5th point is (-1,-2)


-------------- Let's find another point --------------


Start with the given function


Plug in


Raise 0 to the 2nd power to get 0


Multiply 1 and 0 to get 0


Subtract 3 from 0 to get -3


So when , we have


So our 6th point is (0,-3)


-------------- Let's find another point --------------


Start with the given function


Plug in


Raise 1 to the 2nd power to get 1


Multiply 1 and 1 to get 1


Subtract 3 from 1 to get -2


So when , we have


So our 7th point is (1,-2)


-------------- Let's find another point --------------


Start with the given function


Plug in


Raise 2 to the 2nd power to get 4


Multiply 1 and 4 to get 4


Subtract 3 from 4 to get 1


So when , we have


So our 8th point is (2,1)


-------------- Let's find another point --------------


Start with the given function


Plug in


Raise 3 to the 2nd power to get 9


Multiply 1 and 9 to get 9


Subtract 3 from 9 to get 6


So when , we have


So our 9th point is (3,6)


-------------- Let's find another point --------------


Start with the given function


Plug in


Raise 4 to the 2nd power to get 16


Multiply 1 and 16 to get 16


Subtract 3 from 16 to get 13


So when , we have


So our 10th point is (4,13)


-------------- Let's find another point --------------


Start with the given function


Plug in


Raise 5 to the 2nd power to get 25


Multiply 1 and 25 to get 25


Subtract 3 from 25 to get 22


So when , we have


So our 11th point is (5,22)


Now lets make a table of the values we have calculated

 x
y

 -5
22
 
 -4
13
 
 -3
6
 
 -2
1
 
 -1
-2
 
 0
-3
 
 1
-2
 
 2
1
 
 3
6
 
 4
13
 
 5
22
 

Now plot the points. Note: you need to use Firefox to see the next two graphs.



Now connect the points to graph (note: the more points you plot, the easier it is to draw the graph)

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x	f(x)	g(x)	h(x)
-1	-3
		-2
	12

Let's find the first entry of g(x)

Start with the second function


Plug in


Multiply


Subtract


So our table becomes

x	f(x)	g(x)	h(x)
-1	-3	-6
		-2
	12

Let's find the second entry of f(x)


Start with the first function


Plug in


Multiply


Reduce



Now our table updates to

x	f(x)	g(x)	h(x)
-1	-3	-6
	1	-2
	12

Let's find the third entry of x


Start with the first function


Plug in


Divide both sides by 3 to isolate x





Now plug this value into the last row of the x column

x	f(x)	g(x)	h(x)
-1	-3	-6
	1	-2
4	12

Let's find the third entry of g(x)


Start with the second function


Plug in


Multiply


Subtract


Now plug this into the last entry of the g(x) column

x	f(x)	g(x)	h(x)
-1	-3	-6
	1	-2
4	12	9

Now let's find the first entry of h(x)

Start with the given function


Break up the function


Plug in


Plug in , (these values are from the table)


Multiply


Add

x	f(x)	g(x)	h(x)
-1	-3	-6
	1	-2
4	12	9

Now let's find the first entry of h(x)

Start with the given function


Break up the function


Plug in



Plug in , (these values are from the table)



Multiply and reduce


Combine the fractions

x	f(x)	g(x)	h(x)
-1	-3	-6
	1	-2
4	12	9

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Let's denote the first point (9,0) as . In other words, and

Now let's denote the second point (10,6) as . In other words, and



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



Start with the slope formula

Plug in ,,,


Subtract the terms in the numerator to get . Subtract the terms in the denominator to get

Reduce


---------------------
Answer:

So the slope of the line through the points (9,0) and (10,6) is

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Start with the given inequality



Subtract 7x from both sides


Combine like terms on the left side


Divide both sides by -3 to isolate x (note: Remember, dividing both sides by a negative number flips the inequality sign)



Reduce

--------------------------------------------------------------
Answer:
So our answer is (which is approximately in decimal form)




Now let's graph the solution set


Start with the given inequality:



Set up a number line:


Now plot the point on the number line





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


Plug in


Since this inequality is not true, we simply shade the entire portion that does not contain the point x=0 using the point as the boundary. This means we shade everything to the left of the point like this:
Graph of with the shaded region in blue

note: at the point , there is an open circle. This means the point is excluded from the solution set.

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Start with the given expression


Distribute



Group like terms


Combine like terms


So simplifies to


In other words,

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Start with the given inequality



Add 6 to both sides


Combine like terms on the right side

--------------------------------------------------------------
Answer:
So our answer is

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Start with the given equation



Add 10 to both sides


Subtract 3x from both sides


Combine like terms on the left side


Combine like terms on the right side


Divide both sides by 6 to isolate x



Reduce

--------------------------------------------------------------
Answer:
So our answer is (which is approximately in decimal form)

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Start with the original graph





In order to graph the inverse function, simply reflect the original graph over the line (the dashed line)


(note: for some reason, the following image does not display correctly in Internet Explorer. So I recommend the use of Firefox to see this image.)



Graph of the original function (red line) and the inverse function (blue line)

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Start with the given function


To find , simply swap x and f(x) to get




Add 15f(x) to both sides


Subtract x from both sides


Divide both sides by 15 to isolate f(x)


So the inverse function is


Now plug in


Subtract


Divide

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Looking at the graph, notice how the graph extends to infinity in both directions in the x direction. So this shows us that there are no restrictions on the domain.



Since we don't have any restrictions on the domain, this shows us that the domain is all real numbers. In other words, we can plug in any in for x




So the domain of the function in set-builder notation is:





In plain English, this reads: x is the set of all real numbers (In other words, x can be any number)


Also, in interval notation, the domain is:




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




From the graph, we can see that the lowest point is (0,-3). So the smallest that y can be is -3



So the range of the function in set-builder notation is:





In plain English, this reads: y is the set of all real numbers that are greater than or equal to -3

Also, in interval notation, the range is:

[-3,)

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Looking at the graph, notice how the line extends to infinity in all four directions. So this shows us that there are no restrictions on the domain.



Since we don't have any restrictions on the domain, this shows us that the domain is all real numbers. In other words, we can plug in any in for x




So the domain of the function in set-builder notation is:





In plain English, this reads: x is the set of all real numbers (In other words, x can be any number)


Also, in interval notation, the domain is:




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


Also, since there are no restrictions on the range, this means that the range is all real numbers.




So the range of the function in set-builder notation is:





In plain English reads: y is the set of all real numbers (In other words, y can be any number)


Also, in interval notation, the range is:

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Looking at the graph, notice how both the x-values and y-values are restricted. In other words, we cannot plug in x=10 (as it does not fall on the curve). So we can only plug in x-values from -4 to 4. So the domain is


which in set-builder notation is




In plain English, this reads: x is the set of all real numbers that are in between -4 and 4




Also, in interval notation, the domain is:

[-4,4]



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

This also applies to the range. The range is also .



Which in set-builder notation is




In plain English, this reads: y is the set of all real numbers that are in between -4 and 4




Also, in interval notation, the range is:

[-4,4]




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

Summary:

So the domain and range are both [-4,4]

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Looking at , we can see that there are no square roots, logs, and other functions where there are restrictions on the domain.

Also, we can see that the function does not have a division by x (or any combination of variables and constants).
So we don't have to worry about division by zero.


Since we don't have any restrictions on the domain, this shows us that the domain is all real numbers. In other words, we can plug in any in for x




So the domain of the function in set-builder notation is:





In plain English reads: x is the set of all real numbers (In other words, x can be any number)


Also, in interval notation, the domain is:









Now to find the range, simply graph the function to get








From the graph, we can see that the lowest point is (2,0). So the smallest that y can be is 0



So the range of the function in set-builder notation is:





In plain English reads: y is the set of all real numbers that are greater than or equal to 0

Also, in interval notation, the range is:

[0,)

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Looking at , we can see that there are no square roots, logs, and other functions where there are restrictions on the domain.

Also, we can see that the function does not have a division by x (or any combination of variables and constants).
So we don't have to worry about division by zero.


Since we don't have any restrictions on the domain, this shows us that the domain is all real numbers. In other words, we can plug in any in for x




So the domain of the function in set-builder notation is:





In plain English reads: x is the set of all real numbers (In other words, x can be any number)


Also, in interval notation, the domain is:









Now to find the range, simply graph the function to get










From the graph, we can see that the lowest point is (0,-2). So the smallest that y can be is -2



So the range of the function in set-builder notation is:





In plain English reads: y is the set of all real numbers that are greater than or equal to -2

Also, in interval notation, the range is:

[-2,)

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Is this what the function looks like: ?


Start with the given equation


Plug in


Multiply


Subtract


If you have not learned about imaginary numbers, then this is where you would stop. If you have, then...


Simplify the square root





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

OR


Is this what the function looks like: ?


Start with the given equation


Plug in


Multiply


Simplify the square root


Rearrange the terms

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c)

Looking at , we can see that there are no square roots, logs, and other functions where there are restrictions on the domain.

Also, we can see that the function does not have a division by x (or any combination of variables and constants).
So we don't have to worry about division by zero.


Since we don't have any restrictions on the domain, this shows us that the domain is all real numbers. In other words, we can plug in any in for x




So the domain of the function in set-builder notation is:





In plain English reads: x is the set of all real numbers (In other words, x can be any number)


Also, in interval notation, the domain is:









Now to find the range, simply graph the function to get







From the graph, we can see that the lowest point is (0,0). So the smallest that y can be is 0




So the range of the function in set-builder notation is:





In plain English reads: y is the set of all real numbers that are greater than or equal to 0

Also, in interval notation, the range is:

[0,)

---

d)





Looking at , we can see that there are no square roots, logs, and other functions where there are restrictions on the domain.

Also, we can see that the function does not have a division by x (or any combination of variables and constants).
So we don't have to worry about division by zero.


Since we don't have any restrictions on the domain, this shows us that the domain is all real numbers. In other words, we can plug in any in for x




So the domain of the function in set-builder notation is:





In plain English reads: x is the set of all real numbers (In other words, x can be any number)


Also, in interval notation, the domain is:









Now to find the range, simply graph the function to get











From the graph, we can see that the lowest point is (-2,0). So the smallest that y can be is 0



So the range of the function in set-builder notation is:





In plain English reads: y is the set of all real numbers that are greater than or equal to 0

Also, in interval notation, the range is:

[0,)

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a)

Looking at , we can see that there are no square roots, logs, and other functions where there are restrictions on the domain.

Also, we can see that the function does not have a division by x (or any combination of variables and constants).
So we don't have to worry about division by zero.


Since we don't have any restrictions on the domain, this shows us that the domain is all real numbers. In other words, we can plug in any in for x




So the domain of the function in set-builder notation is:





In plain English reads: x is the set of all real numbers (In other words, x can be any number)


Also, in interval notation, the domain is:









Now to find the range, simply graph the function to get







From the graph, we can see that the lowest point is (0,3). So the smallest that y can be is 3




So the range of the function in set-builder notation is:





In plain English reads: y is the set of all real numbers that are greater than 3

Also, in interval notation, the range is:

[3,)

---

b)





Looking at , we can see that there are no square roots, logs, and other functions where there are restrictions on the domain.

Also, we can see that the function does not have a division by x (or any combination of variables and constants).
So we don't have to worry about division by zero.


Since we don't have any restrictions on the domain, this shows us that the domain is all real numbers. In other words, we can plug in any in for x




So the domain of the function in set-builder notation is:





In plain English reads: x is the set of all real numbers (In other words, x can be any number)


Also, in interval notation, the domain is:









Now to find the range, simply graph the function to get











From the graph, we can see that the lowest point is (0,-5). So the smallest that y can be is -5



So the range of the function in set-builder notation is:





In plain English reads: y is the set of all real numbers that are greater than -5

Also, in interval notation, the range is:

[-5,)

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Remember, the number 24,130 can be broken up like this:









So this shows us that which means . Also, which means . This continues until you run out of variables.

So we have the values

, , , and



So becomes

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Start with the given expression


Group like terms


Combine like terms

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Let's verify the solution



Start with the given equation


Plug in . In other words, replace each with



Square and multiply


Combine like terms


Combine the fractions on the right side and reduce


Since the two sides of the equation are equal, this verifies our answer.




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




Let's verify the solution



Start with the given equation


Plug in . In other words, replace each with



Square and multiply


Combine like terms



Notice how there are denominators of zero. Since division by zero is undefined, is not a solution of the equation.




So our only solution is

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# 1

d)

Start with the given function


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.



Subtract 9 from both sides


Combine like terms on the right side





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

So our domain is:

which in plain English reads: x is the set of all real numbers except

So our domain looks like this in interval notation


note: remember, the parenthesis excludes -9 from the domain

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




Now to find the range, simply graph the function to get


note: ignore the vertical lines, they are not part of the graph and are asymptotes



Looking at the graph, we can see that y can be any number except 0. So .




So our range is:

which in plain English reads: y is the set of all real numbers except

So our range looks like this in interval notation


note: remember, the parenthesis excludes 0 from the range

---

e)





Start with the given function


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.




Factor the left side (note: if you need help with factoring, check out this solver)




Now set each factor equal to zero:

or

or Now solve for x in each case


So our solutions are or



Since and make the denominator equal to zero, this means we must exclude and from our domain

So our domain is:

which in plain English reads: x is the set of all real numbers except or

So our domain looks like this in interval notation


note: remember, the parenthesis excludes -2 and 2 from the domain



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




Now to find the range, simply graph the function to get




note: ignore the vertical lines, they are not part of the graph and are asymptotes




Looking at the graph, we can see that y can be any number except 0. So .




So our range is:

which in plain English reads: y is the set of all real numbers except

So our range looks like this in interval notation


note: remember, the parenthesis excludes 0 from the range

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# 1

a)




Looking at , we can see that there are no square roots, logs, and other functions where there are restrictions on the domain.

Also, we can see that the function does not have a division by x (or any combination of variables and constants).
So we don't have to worry about division by zero.


Since we don't have any restrictions on the domain, this shows us that the domain is all real numbers. In other words, we can plug in any in for x




So the domain of the function in set-builder notation is:





In plain English reads: x is the set of all real numbers (In other words, x can be any number)


Also, in interval notation, the domain is:









Now to find the range, simply graph the function to get









From the graph, we can see that the function's y-values extend forever in both directions. So the range is also all real numbers

So the range of the function in set-builder notation is:





In plain English reads: y is the set of all real numbers (In other words, y can be any number)


Also, in interval notation, the range is:

---

b)





Looking at , we can see that there are no square roots, logs, and other functions where there are restrictions on the domain.

Also, we can see that the function does not have a division by x (or any combination of variables and constants).
So we don't have to worry about division by zero.


Since we don't have any restrictions on the domain, this shows us that the domain is all real numbers. In other words, we can plug in any in for x




So the domain of the function in set-builder notation is:





In plain English reads: x is the set of all real numbers (In other words, x can be any number)


Also, in interval notation, the domain is:









Now to find the range, simply graph the function to get










From the graph, we can see that the function's y-values extend forever in both directions. So the range is also all real numbers

So the range of the function in set-builder notation is:





In plain English reads: y is the set of all real numbers (In other words, y can be any number)


Also, in interval notation, the range is:

---

c)





Looking at , we can see that there are no square roots, logs, and other functions where there are restrictions on the domain.

Also, we can see that the function does not have a division by x (or any combination of variables and constants).
So we don't have to worry about division by zero.


Since we don't have any restrictions on the domain, this shows us that the domain is all real numbers. In other words, we can plug in any in for x




So the domain of the function in set-builder notation is:





In plain English reads: x is the set of all real numbers (In other words, x can be any number)


Also, in interval notation, the domain is:









Now to find the range, simply graph the function to get









From the graph, we can see that the function's y-values extend forever in both directions. So the range is also all real numbers





So the range of the function in set-builder notation is:





In plain English reads: y is the set of all real numbers (In other words, y can be any number)


Also, in interval notation, the range is:

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Start with the given system of equations:





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





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


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



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



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




Multiply the top equation (both sides) by
Multiply the bottom equation (both sides) by




Distribute and multiply





Now add the equations together. In order to add 2 equations, group like terms and combine them



Combine like terms and simplify



Notice how the x terms cancel out




Simplify




Divide both sides by to isolate y




Reduce



Now plug this answer into the top equation to solve for x

Start with the first equation



Plug in




Multiply



Multiply both sides by the LCM of 31. This will eliminate the fractions (note: if you need help with finding the LCM, check out this solver)



Distribute and multiply the LCM to each side



Subtract 258 from both sides


Combine like terms on the right side


Divide both sides by 124 to isolate x



Reduce




So our answer is
and



which also looks like

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I'll do the first two to get you started


# 1




Let's denote the first point (4,2) as . In other words, and

Now let's denote the second point (7,8) as . In other words, and



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



Start with the slope formula

Plug in ,,,


Subtract the terms in the numerator to get . Subtract the terms in the denominator to get

Reduce


---------------------
Answer:

So the slope of the line through the points (4,2) and (7,8) is

---

# 2




Let's denote the first point (-7,-9) as . In other words, and

Now let's denote the second point (3,-2) as . In other words, and



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



Start with the slope formula

Plug in ,,,


Subtract the terms in the numerator to get . Subtract the terms in the denominator to get

---------------------
Answer:

So the slope of the line through the points (-7,-9) and (3,-2) is

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# 1


Start with the given equation


Subtract from both sides


Rearrange the equation


Divide both sides by


Break up the fraction


Reduce



So the equation is now in slope-intercept form () where the slope is and the y-intercept is


So to get the equation into function form, simply replace y with f(x)



So the equation changes to the function






Looking at we can see that the equation is in slope-intercept form where the slope is and the y-intercept is


Since this tells us that the y-intercept is .Remember the y-intercept is the point where the graph intersects with the y-axis

So we have one point




Now since the slope is comprised of the "rise" over the "run" this means


Also, because the slope is , this means:




which shows us that the rise is -4 and the run is 1. This means that to go from point to point, we can go down 4 and over 1



So starting at , go down 4 units


and to the right 1 unit to get to the next point



Now draw a line through these points to graph

So this is the graph of through the points and

---

# 2





Start with the given equation


Subtract from both sides


Rearrange the equation


Divide both sides by


Break up the fraction


Reduce



So the equation is now in slope-intercept form () where the slope is and the y-intercept is



So to get the equation into function form, simply replace y with f(x)



So the equation changes to the function





Looking at we can see that the equation is in slope-intercept form where the slope is and the y-intercept is


Since this tells us that the y-intercept is .Remember the y-intercept is the point where the graph intersects with the y-axis

So we have one point




Now since the slope is comprised of the "rise" over the "run" this means


Also, because the slope is , this means:




which shows us that the rise is 2 and the run is 1. This means that to go from point to point, we can go up 2 and over 1



So starting at , go up 2 units


and to the right 1 unit to get to the next point



Now draw a line through these points to graph

So this is the graph of through the points and

You can put this solution on YOUR website!

Notice if we graph , we get






From the graph, we can see that the curve violates the horizontal line test. In other words, if you pass a straightedge horizontally through the graph, the straightedge will touch two points. So the function is not one-to-one.

You can put this solution on YOUR website!

Notice how the graph does not violate the horizontal or vertical line test. So this shows us that the function is one-to-one.