SOLUTION: Y = ax^2 + 4x + c the maximum is at (1,8) the left hand side starts at (-1,0) it goes through (0,6) and reaches the maximum then goes down an end at (3,0) i need to know t

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Y = ax^2 + 4x + c the maximum is at (1,8) the left hand side starts at (-1,0) it goes through (0,6) and reaches the maximum then goes down an end at (3,0) i need to know t      Log On


   

Question 203258: Y = ax^2 + 4x + c
the maximum is at (1,8)
the left hand side starts at (-1,0)
it goes through (0,6) and reaches the maximum then goes down an end at (3,0)
i need to know the value of c and a.
I have been stuck on this question for a long time. thank you again
i dont even know how to get started.
thank you for taking your time to solve this.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
ax%5E2%2B4x%2Bc
Find a and c.
First, you are told that the maximum (vertex) of the parabola is at (1, 8).
The value of the x-coordinate of the vertex is given by:
x+=+%28-b%29%2F2a Substitute b = 4 and x = 1 (from (1, 8)) to get:
1+=+-4%2F2a Multiply both sides by a.
a+=+-4%2F2 Simplify.
highlight%28a+=+-2%29 so we have...
y+=+-2x%5E2%2B4x%2Bc Now to find the value of c, just substitute the x- and y-coordinates of any of the points given in the problem. (1, 8) or (-1, 0) or (0, 6) or (3, 0).
The reason you can do this is because every one of these points lies on the curve (parabola) and thus satisfies the given equation.
Let's choose the point (0, 6).
6+=+-2%280%29%5E2%2B4%280%29%2Bc Simplify.
highlight_green%286+=+c%29
The equation is:
highlight%28y+=+-2x%5E2%2B4x%2B6%29
Let's see what the curve looks like!
graph%28400%2C400%2C-5%2C5%2C-5%2C10%2C-2x%5E2%2B4x%2B6%29