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/2a ±√(b^2-4ac)/2a
Note that axis of symmetry lies midway between the roots, at x = -b/2a
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!!!
1 answer
1 answer