Question 1170151


 {{{f(x)=-2x^2-10x-12}}}

Find the following for this parabola:

A) The vertex:

write it in vertex form by completing square

{{{f(x)=-2(x^2+5x)-12}}}

{{{f(x)=-2(x^2+5x+b^2) -(-2b^2)-12}}}.......{{{b=5/2}}}

{{{f(x)=-2(x^2+5x+(5/2)^2) +2(5/2)^2-12}}}

{{{f(x)=-2(x+5/2)^2 +2(25/4)-12}}}

{{{f(x)=-2(x+5/2)^2 +1/2}}} -> {{{h=-5/2}}} and {{{k=1/2}}}

=> The vertex: at ({{{-5/2}}} ,{{{1/2}}})

B) The vertical intercept is the point:

will be at ({{{0}}} ,{{{y}}})

{{{f(x)=-2x^2-10x-12}}}

{{{f(x)=-2*0^2-10*0-12}}}

{{{f(x)=-12 }}}


C) Find the coordinates of the two x intercepts of the parabola and write them as a list, separated by commas:

will be at ({{{x}}} ,{{{0}}})

{{{0=-2x^2-10x-12}}}...factor
{{{0=-2(x^2+5x+6)}}}
{{{0=-2(x^2+3x+2x+6)}}}
{{{0=-2((x^2+3x)+(2x+6))}}}
{{{0=-2 (x + 3) (x + 2)}}}

=> {{{x=-3}}} and {{{x=-2}}}

solutions: {{{-3}}} ,{{{-2}}}