Question 118943
given:

vertex at ({{{-3}}},{{{4}}})..=> {{{h=-3}}} and {{{k=4}}}

y-intercept at {{{y=1}}}

we will need to use this  formula to find {{{a}}}:

{{{h= -b/2a}}}}



First use vertex form to find {{{b}}}:

{{{y = a(x-h)^2 + k}}}

{{{1= a(x-(-3))^2 + 4}}}

{{{1- 4 = a(x +3)^2 }}}

{{{- 3 = a (x^2 +6x  +9) }}}

{{{a = (x^2 +6x  +9)/(-3) }}}
	
{{{a = (1/3)x^2 – 2x  -3}}}……………………{{{b=-2}}}


Now use the formula  {{{h= -b/2a}}} to find a:

{{{h= -b/2a}}}

{{{2a= -b/h}}}

{{{a= -b/(2h)}}}
	

{{{a= -b/(2(-3))}}}	
{{{a= -(-2)/(2(-3))}}}

{{{a= 2/-6}}}

{{{a= -(1/3)}}}…………… the stretch/compression factor

Here is the graph:

*[invoke completing_the_square "(-1/3) ", -2, 1]