Question 1198644


Given the equation {{{ f(x)=-3(x-4)(x+8)}}}, determine the following algebraically. please show your work.


{{{ x}}}-intercepts- set {{{ f(x)=0}}}

{{{ -3(x-4)(x+8)=0}}}

{{{ x=4}}}

{{{ x=-8}}}

{{{ x}}}-intercepts: ({{{ 4}}},{{{ 0}}}), ({{{ -8}}},{{{ 0}}})


{{{ y}}}-intercept:- set {{{ x=0}}}

{{{ f(0)=-3(0-4)(0+8)=96}}}


Equation of the axis of symmetry

The axis of symmetry equation of a parabola whose vertex is ({{{ h}}},{{{  k}}}) and opens up/down is {{{ x = h}}}.

so, first convert  to vertex form (3.)

 {{{ x =-2}}}

Vertex:({{{ h}}}, {{{ k}}}) =({{{ -2}}},{{{  108}}}) 

3.Convert the following from standard form to vertex form by completing the square.

{{{ f(x)=-3(x-4)(x+8)}}}

{{{ f(x)=-3 x^2 - 12x + 96}}}

{{{ f(x)=-3 (x^2 + 4x )+ 96}}}

{{{ f(x)=-3 (x^2 + 4x +2^2)-(-3)*2^2+ 96}}}

 {{{ f(x)=-3 (x +2)^2)+12+ 96}}}

 {{{ f(x)=-3 (x +2)^2)+108}}}

so {{{ h=-2}}}, {{{ k=108}}}