Answer:
x = -1 <--- axis of symmetry
(-1, -9) <-- vertex
2, -4 (or, in my old precalc class we wrote our roots as (2,0) and (-4,0)) <--- roots
Step-by-step explanation:
a) Axis of symmetry equation is defined as
[tex]-b/2a[/tex] = x (i memorized this)
Quadratic equation form:
[tex]ax^2+bx+c[/tex]
And our original equation:
[tex]x^2+2x-8[/tex]
So..
[tex]-2/2 = - 1[/tex]
So our axis of symmetry is at x = -1,
b) now vertex point can be found by plugging this value back into our original equation.
y = [tex](-1)^2+2(-1)-8[/tex]
y = 1 - 2 - 8
y = - 9
So our vertex is at (-1, -9)
c) Finally, lets factor
[tex]x^2+2x-8[/tex] = 0 (when finding the roots, set y = 0, because we are trying to find where we cross the x-axis)
[tex]x^2-2x+4x-8[/tex] = 0
[tex]x(x-2) + 4(x-2) = 0[/tex]
[tex](x-2)(x+4)=0[/tex]
x = 2; x = -4 <--- Roots
