Question 1002163

The parabola {{{y=ax^2+k}}} has vertex ({{{0}}},{{{7}}}) and passes through the point ({{{3}}},{{{2}}}). 

Find its equation.

use vertex form of parabola:

{{{y=a(x-h)^2+k}}}....if vertex ({{{0}}},{{{7}}}), we have

{{{y=a(x-0)^2+7}}}

{{{y=ax^2+7}}}......use point ({{{3}}},{{{2}}}) to find {{{a}}}

{{{2=a*3^2+7}}}

{{{2=9a+7}}}

{{{2-7=9a}}}

{{{-5=9a}}}

{{{a=-5/9}}}

so, your equation is: {{{y=-(5/9)x^2+7}}}


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(0,7,.12),circle(3,2,.12),
locate(0.3,7,V(0,7)),locate(3,2,p(3,2)),
 graph( 600, 600, -10, 10, -10, 10, -(5/9)x^2+7)) }}}