Question 687718: Find the equation of the parabola that passes through the points (2,3) and (10,3) and has a max value of y=35.
Can you please help me out? thanks in advance
Found 2 solutions by MRperkins, ankor@dixie-net.com:
Answer by MRperkins(300)   (Show Source):
You can put this solution on YOUR website! email me and I will work this one out with you in an online whiteboard.
http://www.scribblar.com/mg9yqg9
Basically, we are dealing with a quadratic function. Generally, that will involve using the "vertex form"
so, we know that there is a maximum of y=35. what, in the formula, is the y value of the vertex of the parabola? (b)
Since the y values are the same on the given points, then the axis of symmetry is in the middle of them (2+10)/2 or 6. that is the h value of the vertex. If they were not the same y values, then we could still find the h value, but it would be more complicated:
.
.
***alternative way to find h value Start***
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so we can insert that into the formula: y=a(x-h)^2+35
we also know two points that work. enter the first point and you get 3=a(2-h)^2+35
foil (2-h)^2 to get h^2-4h+4,
solve for "a"
a=-32/(h^2-4h+4)
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now plug the other point into y=a(x-h)^2+35
and you get 3=a(10-h)^2+35
subtract 35 from both sides
-32=a(10-h)^2
substitute -32/(h^2-4h+4) from the first equation in for a in the second equation.
You get:
-32=(-32(10-h)^2)/(h^2-4h+4)
reduce the -32's and move the h^2-4h+4 to the left
reduce to find the h value.
h=6
***alternative way to find h value Stop***
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so now you have y=a(x-6)^2+35
use one of the points for the x and y value to find "a"
3=a(4)^2+35
-32/16=a
a=-2
so
y=-2(x-6)^2+35
Answer by ankor@dixie-net.com(22740)  (Show Source):
You can put this solution on YOUR website! Find the equation of the parabola that passes through the points (2,3) and (10,3) and has a max value of 35
:
From the two points, 2,3 and 10,3, we know the axis of symmetry is halfway between them and is the x coordinate of the max, so we have a third point, 6,35
:
Using the y = ax^2 + bx + c form we can solve for a, b, c; three equations:
x=2, y=3
4a + 2b + c = 3
:
x=6, y=35
36a + 6b + c = 35
:
x=10, y=3
100a + 10b + c = 35
:
Eliminate c
36a + 6b + c = 35
4a + 2b + c = 3
=------------------subtraction eliminates c
32a + 4b = 32
Simplify, divide by 4
8a + b = 8
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Eliminate c again
100a + 10b + c = 3
36a + 6b + c = 35
----------------------
64a + 4b = -32
Simplify, divide by 4
16a + b = -8
:
Use these two equations to eliminate b
16a + b = -8
8a + b = 8
---------------subtracting eliminates b, find a
8a = -16
a = -16/8
a = -2
:
Find b using 8a + b = -8
8(-2) + b = 8
b = 8 + 16
b = 24
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Find c using 4a + 2b + c = 3
4(-2) + 2(24) + c = 3
-8 + 48 + c = 3
c = 3 - 40
c = -37
:
The equation: y = -2x^2 + 24x -37
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