Question 66247This question is from textbook
: Please help me solve this equation. "Find a, b, and c so that the parabola whose equation is y=ax^2+bx+c has its vertex (3,2) and passes through the point (-1,10)". I have tried plugging in the point, but I cannot seem to figure it out. Thank you.
This question is from textbook
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Please help me solve this equation. "Find a, b, and c so that the parabola whose equation is y=ax^2+bx+c has its vertex (3,2) and passes through the point (-1,10)". I have tried plugging in the point, but I cannot seem to figure it out.
:
Two coordinates are given: the vertex x=3,y=2 and x=-1,y=10
:
Using x=3,y=2, the 1st equation
a(3^2) + 3b + c = 2
9a + 3b + c = 2
:
Using x=-1,y=10, find the 2nd equation
a(-1^2) - 1b + c = 10
a - b + c = 10
:
They gave x=3,y=2 as the vertex so the axis of symmetry would be x=3 and from that we can derive a 3rd coordinate.
The point x=-1 is 4 units to the left of +3 so if we go 4 units to the right of
+3 we have a point x=7,y=10.
The 3rd equation would be a(7^2) + 7b + c = 10; which is: 49a + 7b + c = 10
:
We have 3 equations and 3 unknowns a,b,c
Subtract the 1st eq from the 3rd eq, eliminate c:
49a + 7b + c = 10
9a + 3b + c = 2
--------------------subtract
40a + 4b = 8
:
Subtract the 2nd equation from the 1st equation, eliminate c again:
9a + 3b + c = 2
a - b + c = 10
-----------------Subtract
8a + 4b = -8
:
Subtract the resulting equation and eliminate b
40a + 4b = 8
8a + 4b = -8
--------------Subtract
32a = +16
a = 16/32
a = .5
:
Use eq, 8a + 4b = -8, substitute .5 for a to find b:
8(.5) + 4b = -8
4 + 4b = -8
4b = -12
b = -12/4
b = -3
:
Substitute in the 2nd equation to find c:
.5 - (-3) + c = 10
3.5 + c = 10
c = 10-3.5
c = +6.5
:
our equation is: y = .5x^2 - 3x + 6.5
It would look like this:
:
Note that it passes thru coordinates -1,10 and vertex is 3,2
:
Could you follow this OK?
|
|
|
