Question 194574
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Given: f(x)=2x^2 +8x-42

Simplified by dividing the whole eqn by 2:
{{{cross(2)x^2/cross(2)+cross(8)4*x/cross(2)-cross(42)21/cross(2)=0}}}
{{{x^2+4x-21=0}}}, perfect square
So, {{{(x+7)(x-3)=0}}}
Then, {{{red(x=-7)}}}; {{{red(x=3)}}}, X-Intercepts.


To follow up on this, let us find the vertex: {{{a(x-h)^2+k}}}


Complete the square:
{{{(x^2+4x+highlight(4))-21-highlight(4)}}}
{{{(x+2)^2-25}}}


Vertex: (-2,-25)


Since a>0, (a=1), the parabola opens upwards.

{{{drawing(400,400,-10,10,-30,5,grid(1),graph(400,400,-10,10,-30,5,x^2+4x-21),blue(circle(-2,-25,.18)),blue(circle(-7,0,.18)),blue(circle(3,0,.18)))}}}


Thank you,
Jojo</font>