Question 993634


given:the vertex is ({{{3}}},{{{-11}}}) and the y-intercept is {{{-2}}}

use vertex form of equation:

{{{y = a(x - h)^2 + k}}} where vertex: ({{{h}}},{{{k}}}) and y-intercept: ({{{x}}},{{{y}}})

so, we will use given points to find coefficient {{{a}}}:

vertex: ({{{h}}},{{{k}}})= ({{{3}}},{{{-11}}}) and 
the y-intercept ({{{x}}},{{{y}}})= ({{{0}}},{{{-2}}})

{{{-2 = a(0 - 3)^2 -11}}} 

{{{-2 = a( - 3)^2 -11}}}

{{{-2+11 = 9a }}} 

{{{9 = 9a }}} 

{{{a=1}}}

so your equation is:

{{{y = (x - 3)^2 -11}}}


{{{drawing( 600, 600, -15, 15, -15, 15,
circle(3,-11,.2),locate(3,-11,V),circle(0,-2,.2),
 graph( 600, 600, -15, 15, -15, 15, (x - 3)^2 -11)) }}}