Question

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

Answer 1

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

$-b/2a$ = x  (i memorized this)

Quadratic equation form:

$ax^2+bx+c$

And our original equation:

$x^2+2x-8$

So..

$-2/2 = - 1$

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 = $(-1)^2+2(-1)-8$

y = 1 - 2 - 8

y = - 9

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

c) Finally, lets factor

$x^2+2x-8$ = 0  (when finding the roots, set y = 0, because we are trying to find where we cross the x-axis)

$x^2-2x+4x-8$ = 0

$x(x-2) + 4(x-2) = 0$

$(x-2)(x+4)=0$

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