Question

find the vertex and axis of symmetry.
f(x) = -x^2 +6x+ 6

Answer 1

A formula for the axis of symmetry can be found by simply averaging the roots from the quadratic formula:\[x=\frac{1}{2}\left[\frac{-b+\sqrt{b^2-4ac}}{2a}+\frac{-b-\sqrt{b^2-4ac}}{2a}\right]=\boxed{-\dfrac{b}{2a}}\]The vertex is then where this axis of symmetry intersects the parabola.

Answer 2

that's awesome yakeyglee ^^^

Answer 3

Thanks!

Answer 4

since you know where the axis of symmetry is, you can use that to find the vertex...
HINT: the vertex has the same x coordinate as the line of symmetry.

Answer 5

For any parabola of the form \(ax^2+bx+c=0 \) the vertex will be at

\( \large \left(-\frac b {2a} , -\frac {b^2-4ac}{4a}\right) \).

Equation of line of symmetry will be at \(x=-\frac b {2a} \)