Y=x^{2} +2x-8 determine the axis of symmetry, the vertex point and find the roots by factoring

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

= x  (i memorized this)

Quadratic equation form:

And our original equation:

So..

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 =

y = 1 - 2 - 8

y = - 9

So our vertex is at (-1, -9)

c) Finally, lets factor

= 0  (when finding the roots, set y = 0, because we are trying to find where we cross the x-axis)

= 0

x = 2; x = -4 <--- Roots

72 is the answer to this problem