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

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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

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

User Lelly
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