The graph of the function f(x) = ax^2 + bx + c has a vertex at x = 1 and passes through the points (0, 1) and (-1, -8). Find a, b and c.

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- Jan 24th 2007, 12:48 PMsymmetryFind a, b and c
The graph of the function f(x) = ax^2 + bx + c has a vertex at x = 1 and passes through the points (0, 1) and (-1, -8). Find a, b and c.

- Jan 24th 2007, 10:35 PMearboth
Hello,

to calculate 3 values you need 3 equations. You know that the x-value of the vertex is $\displaystyle -\frac{b}{2a}$. So you have:

$\displaystyle -\frac{b}{2a}=1 \Longleftrightarrow b = -2a$

The coordinates of the given points must satisfy the equation of the function. Plug in the coordinates:

$\displaystyle 1=a \cdot 0^2+b \cdot 0 + c \Longleftrightarrow c=1$

$\displaystyle -8=a \cdot (-1)^2+b \cdot (-1) + c $

Solve for a, b and c. Use the method which is the most convenient for you. I've got:

a = -3, b = 6 and c = 1

EB - Jan 25th 2007, 02:58 PMsymmetryok
I thank you for your help and insight.