Question

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

Answer 1

have you covered parabolas yet?

Answer 2

no i have not

Answer 3

well, you may want to cover that first I'd think

Answer 4

@malik101: By "turning point," do you by any chance mean "vertex?"

Answer 5

Otherwise...what would "turning point" mean?

Answer 6

The general parabola \(ax^2+bx+c\) has a vertex at \(\left(-\dfrac{b}{2a},~-\dfrac{b^2-4ac}{4a}\right)\).

Given that the turning point/vertex is (0,2), you have the equations:
\[\begin{cases}-\dfrac{b}{2a}=0\\\\
-\dfrac{b^2-4ac}{4a}=2\end{cases}\]
You have enough info here to solve for some but not all of the coefficients.

To solve for the remaining unknown, plug in the other known point, (1,8):
\[\begin{align*}f(1)&=a(1)^2+b(1)+c\\8&=a+b+c\end{align*}\]