Question 1204077

{{{ -12(x+3)=(y-3)^2}}} 
Focus = ({{{ -6}}} ,{{{ 3}}} )
Directrix = {{{ x=0}}} 
Vertex = ({{{ -3}}} ,{{{ 3}}} )

Two points determine any line. However, since a parabola is curved, we should find more than two points. In this text, we will determine at least five points as a means to produce an acceptable sketch. 

To begin, we graph our first parabola by plotting points. Given a quadratic equation of the form {{{  -12(x+3)=(y-3)^2}}} ,{{{  x}}}  is the independent variable and {{{ y}}}  is the dependent variable (so, you choose it). Choose some values for {{{ x}}}  and then determine the corresponding {{{ y}}} -values. Then plot the points and sketch the graph.

since you are already given parabola opening sideways and focus is at ({{{ -6}}} ,{{{ 3}}} ), means the parabola is opening to the {{{ left}}}  and graph will be in II and III quadrant
you are also given two points, use them and use formula to find three more

{{{ -12(x+3)=(y-3)^2}}} ...let {{{ x=-4}}} 
{{{ -12(-4+3)=(y-3)^2}}} 
{{{ y=6.46 }}}   or {{{y=-0.46}}} 

{{{ -12(x+3)=(y-3)^2}}} ...let {{{ x=-10}}} 
{{{ -12(-10+3)=(y-3)^2}}} 
{{{ y=12.16}}}   or {{{-6.16}}} 

{{{x }}} |{{{y}}} 
{{{-6}}} |{{{3}}} 
{{{-3}}} |{{{3}}} 
{{{-4}}} |{{{6.46}}} 
{{{-4}}} |{{{-0.46}}} 
{{{-10}}} |{{{12.16}}} 
{{{-10}}} |{{{-6.16}}} 


plot these points and sketch the graph



{{{ drawing( 600, 600, -20, 10, -20,20,
locate(-6,3,F(-6,3)),circle(-6,3,.15),circle(-3,3,.15),
green(line(-0.2,-20,-0.2,20)),locate(-3,3,V(-6,3)),
circle(-4,6.46,.15),circle(-4,-0.46,.15),
circle(-10,12.16,.15),circle(-10,-6.16,.15),
graph( 600, 600, -20, 10, -20, 20, 3 -2sqrt(3)*sqrt(-x-3),2sqrt(3)*sqrt(-x-3)+3)) }}}