Question

For the quadratic function f(x) = - x^2 +2x + 6, find the vertex and the axis of symmetry, and graph the function.

Again, I need help with this as these are practice examples for an upcoming quiz and I stink at this part of algebra and I need to get it otherwise I won't graduate and I have worked really hard at this, so please I need help with knowing how to do this, Please!!!

Answer 1

for F(x) = ax^2+bx+c, the vertex lies on the axis of symmetry, at x = -b/2a.

so, for your function, that is at

x = -2/-2 = 1
f(1) = 7

so, the vertex is at (1,7). See the graph at

http://www.wolframalpha.com/input/?i=-+x^2+%2B2x+%2B+6

Recall that the quadratic formula says that the roots of such a polynomial lie at

x = [-b±√(b^2-4ac)]/(2a)

That is just

x = <b>-b/2a</b> ±√(b^2-4ac)/2a

Note that axis of symmetry lies midway between the roots, at x = -b/2a