# Questions on Algebra: Coordinate systems, graph plotting, etc answered by real tutors!

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Question 994328: let f(x)=(-1/2)X^2-3X+(5/2)
a. y intercept
b. x intercept
c. axis of symmetry
d. vertex
f. sketch
THANK YOU!!!

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So the vertex is (,) and the axis of symmetry is .
To get the y-intercept, set .

(,)
To get the x-intercepts, set .

.
.
.
.

Question 994575: what is the vertex of y=-(x+2)2squared + 10 ?
Found 2 solutions by MathTherapy, josgarithmetic:
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what is the vertex of y=-(x+2)2squared + 10 ?
Your equation is in vertex form, and since the vertex form of a quadratic is: , it's clear that the vertex, (h, k) is: ()



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Write it this way:
y=-(x+2)^2+10

The rendering tags will make it look like it should:

This is in standard form. Read the vertex from the equation.
(-2,10)

Question 994534: I can not figure this problem out can someone help pls
Natalie performs a chemistry experiment where she records the temperature of an ongoing reaction. The solution is 93.5º C after 3 minutes; 90º C after 5 minutes, 84.8 C after 9 minutes; 70.2º C after 18 minute; 54.4º C after 30 minutes; 42.5ºC after 37 minutes; and 24.9º C after 48 minutes. Perform a linear regression on this data to complete the following items.
Write the equation of the best-fitting line.

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plot your points, let time be and temperature
|º
|º
|º
|º
|º
|º
|º
|º

To calculate the slope, pick two points on the line, and calculate the slope as:
(your slope should have a negative value).
The equation of the line is then
where
I would pick these two points on the line:
(,)
(,)

so, equation is
use one point to find

Question 994417: How do I graph the vertex and axis of : f(x)=(x+3)
plus it said something about symmetry?????

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No vertex and no axis of symmetry. f(x) is a linear function, slope 1 and y-axis intercept 3.

Question 992589: Find a formula that expresses the fact that an arbitrary point P(x, y) is on the perpendicular bisector ℓ of segment AB.
A(−4, 7), B(8, −13)

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Any pt. on perpendicular bisector of a line segment is equidistant from both the end points.
So, Distance from P to A = Distance from P to B
=>
Squaring both sides, & simplifying,

=>
=> represents all such points P as required.

Question 993809: The points (5,3) and (14,-8) form the endpoints of a diameter of a circle in the xy-plane. Find the x-intercepts of the circle. Please help, a step-by-step explanation would be appreciated as I need to ensure that I understand how to solve for these types of problems. Thank you!
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Standard Form equation for a circle is . The midpoint of the two given points is the center of the circle. This center will be the point (h,k). The distance using the Distance Formula between (h,k) and either of the two given endpoints gives you your radius, r.

Question 993810: The points (2,7) and (8,3) are equidistant to a point P on the y-axis. Find this point P. Step-by-step explantions are greatly appreciated so I can throughly understand how to solve for these types of problems. Thank you in advance!
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The point P on the y-axis would be some ordered pair (0,y) and you want to solve for y. The initial equation uses The Distance Formula.

The points (2,7) and (8,3) are equidistant to a point P on the y-axis.

Simplify and solve for y.

Question 993811: Find an equation for the line that is tangent to the circle x^2+y^2=169 at the point (5,12).
I need help beind reminded of the equation of a circle and equations related to this, and also a step-by-step explantion to help me to understand how to solve this typ of problems and others efficiently. Thank you so much!

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Center of circle on the Origin, , and for this, radius is r.
You have , so the radius .

The point ON THE CIRCLE, (5,12), is part of a tangent line which passes through this point. This means, you want to find an equation for this line and this line TOUCHES the circle at this point; and it is perpendicular to the line which contains this point (5,12) and the Origin (which is center of your circle.)
-
Work to understand that discussion before continuing.

What is the line containing the circle's center (0,0) and the given point (5,12)? You should find just intuitively this is . The y-intercept and the x-intercept both 0.

What is the equation for the line PERPENDICULAR to and contains the point (5,12)? For perpendicularity, its slope must be negative reciprocal of , so this slope needed will be . You can use the point-slope equation form (for convenience if you are comfortable with it), and plug-in the needed slope and included point (5,12):

Use simple algebra if you want this equation in standard form or in slope-intercept form.

--
Try to make a sketch or a graph on your own to help analyze the problem description.

Question 993691: If the points (2,5) are shifted 5 units up and 4 units down what are the new coordinates

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Not sure you typed the problem correctly, but if you did
Shifting the point up increases the y-coordinate and shifting it
down decreases the y-coordinate, thus
(2,5) becomes (2,10) and then (2,6).

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The equation is in slope-intercept form and represents a line. The y-intercept is (0,8) and the slope is 5, meaning vertical change is 5 units for every 1 unit of horizontal change.

Question 993428: Show that the points A (-1,-1),B (-3,4), and C (4,1) are the vertices of a right triangle.
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Hi there,
AB = y2 - y1/x2 - x1
AB = 4-(-1)/-3-(-1)
AB = 5/-2
AB = -5/2
BC = y2 - y1/x2 - x1
BC = 1 - 4/4-(-3)
BC = -3/7
AC = y2 - y1/x2 - x1
AC = 1 -(-1)/4-(-1)
AC = 2/5
Lines that are perpendicular
that multiply together to give -1
Side AB and side AC's gradients
multiply together to give -1
-5/2 x 2/5 = -1
Right angle is at A
Hope this helps :-)

Question 992590: Find a formula that expresses the fact that P(x, y) is a distance 7 from the origin.

Describe the set of all such points.
This is a circle of radius
, with a center at
(x, y) =

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x^2 + y^2 = 49

here's the graph of that equation.

the vertex form of the eqaution of a circle is:

(x-h)^2 + (y-k)^2 = r^2

r is the radius of the circle..
(h,k) is the (x,y) coordinate of the center of the circle.

in your circle, the center is (0,0) so h = 0 and k = 0

any point p(x,y) that lies on the circle, will be r units away from the center of the circle.

https://www.mathsisfun.com/algebra/circle-equations.html

they use (a,b) as the center of the circle,
i use (h,k) as the center of the circle.
doesn't matter.
they reference the same thing.

a or h references the x-coordinate of the center of the circle.
b or k references the y-coordinate of the center of the circle.

Question 990836: The lines x-y-2=0 and 2x-5y-7=0 intersect at point P. Find co-ordinates of point P. The line through P with gradient 2 meets the axis A and B. Calculate the area of triangle AOB.
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x-y-2=0 ---(i)
2x-5y-7=0---- (ii)
Multiplying eqn (i) by 2,
2x-2y-4=0 ---(iii)
Subtracting (ii) from (iii), we get,
3y+3=0
=> y = -1
Putting back, x= y+2 = 1
So, intersection pt. P is (1,-1)
Let the line through P with gradient 2 be given by,
y = 2x + c
Since it passes through P(1,-1)
-1 = 2*1 + c
=> c = -3
So the required line is
y = 2x - 3
For finding intersection with axes, putting x and y =0 separately, we get, the intersection points as A(0,-3) and B(3/2,0)
So, area of triangle OAB
= (1/2)*(3)*(3/2)
= 9/4 sq. units.

Question 992593: Find the point with coordinates of the form (3a, a) that is in the third quadrant and is a distance 5 from
P(1, 2).
(x, y) =

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=>
=>
=>
Solve for a, and choose only negative value(a is approx -0.618), since the pt. is in 3rd quadrant.

Question 992728: Graph each equation using the gradient and y-intercept for:
y= 5/2x -3

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First plot the y-intercept.
(0,-3)
Slope is equal to change in y over change in x.
So for a change in x of 2, y changes 5 units.
(0,-3)+(2,5)=(2,2)
Plot that point and draw the line.
.

Question 992588: Find all points on the y-axis that are a distance 7 from P(5, 9).
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.
Point(,)=(,)
Point(.)=(,)
d=distance=7
.

.
 Solved by pluggable solver: SOLVE quadratic equation with variable Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=96 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 13.8989794855664, 4.10102051443364. Here's your graph:

.
y=4.1 or 13.9
.
ANSWER: There are two points: (0,4.1) and (0,13.9)
.

Question 992391: FIND FIVE ORDERED PAIRS TO THE EQUATION Y=3x(-1)-1
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Sorry, we can't understand you when you shout. Turn off caps lock and re-post.

John

My calculator said it, I believe it, that settles it

Question 992219: x^2+y^2+6x-6y-107=0 What is the Radius and Center of a circle

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x^2+y^2+6x-6y-107=0 What is the Radius and Center of a circle
---
x^2+y^2+6x-6y = 107
Complete the squares for x and y:

----
--> center at (-3,3)

Question 992124: find the equation which is parallel and perpendicular to 2y+3x-4=0

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.
Straight line  4y + 6x - 4 = 0  is parallel to the original straight line  2y + 3x - 4 = 0.

Straight line  2x - 3y = 0  is perpendicular to the original straight line  2y + 3x - 4 = 0.

Question 991586: If f(−4)=−1, write an ordered pair that must be on the graph of y=f(x−5)+2
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.
This ordered pair is

(x,y) = (1, 1).

I.e.  x=1  and  y=1.

Indeed,  if you substitute  x=1  into  f(x-5),  you will get  f(1-5) = f(-4),  which is  -1  by the condition.

Hence,  f(x-5)+2 = -1 + 2 = 1,  which is the value of  y  in the pair.

Question 991540: An airplane travels 3035km against the wind in 5 hours and 3935km with the wind in the same amount of time. What is the rate of the plane in still air and what is the rate of the wind?
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Same problem, different numbers.
==============================================
an airplane, flying with a tail wind, travels 2000 miles in five hours. The return trip, against the wind, takes eight hours. Find the cruising speed of the plane and the speed of the wind (assuming that both rates are constant)
==================
find the 2 groundspeeds, downwind and upwind.
The plane's airspeed is the average of the 2.
----
Windspeed = difference between airspeed and groundspeed.

Question 989964: For what value of m will the triangle formed by the lines y=5, y=mx-6 and
y=-mx-6 be equilateral?

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The two lines have y-intercepts equal to .
So the altitude of the equilateral triangle is
The altitude, half side, and side form a right triangle.

The top edge is centered about so the endpoints of the two vertices are
(,) and (,).
Now that you have two points, find the slope.

.

Question 989944: Use given endpoint "r" and midpoint "m".find s in rs segment. R=6.-2. M=5.3
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Midpoint is the average of x and y values.

.
.

.
.
S:(4,8)

Question 990981: Express this in general form: Slope: -2/7 containing the point (4,7)
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Equation may begin through point-slope form: . Use your algebra skill to put into a form, .

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==============
The graph of -6x + 3y = 12 is a line of infinite length.
Every point on it has an (x,y) pair.
----
-6x + 3y = 12
y = 2x + 4
---
Pick any value of x and find y.
eg,
x = 0 --> y = 4 --> (0,4)
x = 1 --> y = 6 --> (1,6)
etc

Question 990626: explain in detail please: find all points having an x coordinate of 2 whose distance from the point (-2,-1) is 5.
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(2,y) to be 5 units from (-2,-1).

THIS is the use of the Distance Formula:

These are the possible sequence of arithmetic steps to solve the equation for y.

y=-4 or y=2

Question 990623: hello, this is my homework and I couldn't find the right answer for it. I will provide the step I did to try and solve the problem. Also, I would really appreciate if you could point out the mistake I did so I could improve.

Question:
Find the point (x,y) on the line y=−2x−3 that is equidistant from the points (−1,6) and (−2,9).
my step:
d² = (x-a)² + (x-b)²

= (x-[-1])² + (y-6)²
= x² +2x+1+y²-12y+36=x²+4x+4+y²-18y+81
2x+1-12y+36-4x-4+18y-81=0
-2x+6y-48=0
-2x+6(-2x-3)-48
-2x-12x-18-48
-14x-66
x=-66/14
y=-2x-3
= -2(66/14)-3
= -174/14

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You have an unknown line with general points, (x, -2x-3). This point is same distance from point (-1,6) as from point (-2,9).

Distance Formula will give the equation to express that reworded description.

Analyze that for it to all make sense. You can then continue with the algebraic steps to simplify the equation and solve for x.

Question 990503: which of the following combinations represents the vertex and two x-intercepts of the function given below? y=x^2-2x-35
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Convert to vertex form,

So the vertex is (1,-36) and

and are the x-interecepts.

Question 988854: Find the coordinates of the other endpoint of the segment, given it's midpoint M and one endpoint Q.
M(a,k), Q(f,q)
The second endpoint is P

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Let the missing coordinates of the endpoint P be (x,y),
Mid point given :M(a,k) henceforth
a=(x+f)/2,2a-f
K=(y+q)/2,2k-q
The end point P(2a-f,2k-q)

Question 988589: HI, this question is from my textbook.
The equations of two sides of a square are y=3x-1 and x+3y-6=0. (0,-1) is one vertex of the square. Find the coordinates of the other vertices.

Found 2 solutions by KMST, Alan3354:
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, and are the equations of two sides of the square.
The coordinates of the vertex at the intersection of those two sides can be found by solving the system
--->--->--->--->--->--->---> .
So, point (9/10,17/10), or (0.9,1.7) is one vertex,
the one at the intersection of the two given sides.
Point (0,-1) , with is the other end/vertex of that side with equation .
So far we have this:

What we do not know is to what side of segment to draw the square.
There are really two right answers. Which one will be the right one according to your textbook? Or did the authors realize that there were two possible answers?
We can draw the square to the right, like this:
, or we can draw the square to the left, as Alan suggested.
Either way, if we encase that square inside a larger square with sides parallel to the x- and y-axes,
we form four congruent right triangles with legs measuring
and , like this
So, , ,
, and .

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The equations of two sides of a square are y=3x-1 and x+3y-6=0. (0,-1) is one vertex of the square. Find the coordinates of the other vertices.
---------------
(0,-1) is point A.
Find the intersection of the lines, that's a 2nd vertex, point B.
---
y=3x-1
x+3y-6=0 --> y = (-1/3)x + 2
-----------
x + 3(3x-1) = 6
10x = 9
x = 0.9, y = 1.7 --> B(0.9,1.7)
-----------
The distance between the 2 points is the side length.

-----------
Find the line thru A(0,-1) parallel to x+3y-6=0
--> y = (-1/3)x - 1
A vertex (point D) is on this line distance s from point A.
The x-distance from A to D is 2.7, and the y-distance is 0.9
--> D (-2.7,-0.1)
------------------
The side CD goes thru D with a slope of 3 (parallel to AB)
y + 0.1 = 3(x + 2.7)
y = 3x + 8
---
Vertex C is the intersection of y = 3x + 8 and y = (-1/3)x + 2
3x + 8 = -x/3 + 2
10x/3 = -6
x = -1.8, y = 2.6
C(-1.8,2.6)

Question 988580: To make 14 gallons of orange​ paint, a worker mixes 10 gallons of red paint and 4 gallons of yellow​ paint, but the color is incorrect. What is the correct​ combination? What is the likely error that the worker​ made?
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Either wrong recipe (formula) or mistake in using the formula's quantities. Not much more can reasonably be in this response. Anything more depends on having oneself in the practical situation, and having a color reference sample.

Question 987277: Write an equation in standard form for a line with an x intercept of 3 and y intercept of 6
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Write an equation in standard form for a line with an x intercept of 3 and y intercept of 6
===============
6x + 3y = 18
2x + y = 6

Question 987218: On a graph, there is a point plotted on (-1,2) and (3,0).
What is the:
A) Domain
B) Range
C) C) X values of all zeroes
D) Equations of all asymptotes
E) 1 to 1 or not
F) Continuous or not (if continuous tell where)
G) Increasing, decreasing, both (if both tell where), or neither
H) Bounded above, bounded below, bounded, or unbounded (if unbounded, tell where)

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Does the function consist only of the two given points, or are these two points on a straight line that represents the function? Or are we just supposed to guess what you are talking about. By the way, which ONE of your myriad of questions to you want answered?

John

My calculator said it, I believe it, that settles it

Question 987210: Draw the graph of:
x/6 - y/4 = 1

Found 2 solutions by stanbon, macston:
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Draw the graph of:
x/6 - y/4 = 1
---
Let x = 0, then y = -4
Let y = 0, then x = 6
----
Plot (0,-4) and (6,0) and draw a line thru them
------

Cheers,
Stan H.
--------------

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.
Find x-intercept:
Set y=0

x-intercept is (6,0)
.
Find y-intercept:
Set x=0

y-intercept is (0,-4)

Question 987164: Jane goes for a walk.On the map it shows that she walked 5cm.Calculate the actual distance in kilometres she has walked if the scale of the map is 1:15 000.
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Hi there,
1:15000
Multiply 5 x 15000
5:75000
75000/100 = 750metres
Hope this helps :-)

Question 986437: what is the equation of the Vertical line that passes through the points (6,-5) and (6,4)
What is the equation of the Horizontal line that passes through the points (-5,6) and (4,6)

Question 986230: Y=-2x+3
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What is your question? Not enough info, sorry.

Question 985744: Write an equation in slope-intercept form
Perpendicular to y= 3x - 2
Passes through (6, -1)

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Perpendicular to y= 3x - 2

The equation would be something as .

Passes through (6, -1)

-

Line you want is .

Question 985204: How do you write an equation in standard form that goes through the point (-4,3)
and with a y-intercept of 0?

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given:
the line passes through the point (,)
and with a y-intercept of ; means the line will pass through origin (,)
so, firs we can find the equation of this line in slope-intercept form
 Solved by pluggable solver: Find the equation of line going through points hahaWe are trying to find equation of form y=ax+b, where a is slope, and b is intercept, which passes through points (x1, y1) = (-4, 3) and (x2, y2) = (0, 0). Slope a is . Intercept is found from equation , or . From that, intercept b is , or . y=(-0.75)x + (0) Your graph:

so, the equation of this line in slope-intercept form is
the standard form for linear equations:

so, we have

Question 985086: x=7-y what are the ordered pairs

Question 985085: x=7-y what are the ordered pairs
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x=7-y what are the ordered pairs
------------
Pick values for y and find x.
eg,
y = 0, x = 7 --> (7,0)
y = 7, x = 0 --> (0,7)
y = 100, x = -93 --> (-93,100)
etc

Question 984742: Find all points on the x-axis that are 6 units from the point (4,-3).