Question 1105272
So in the middle where {{{x=0}}}, {{{y=27}}}.
At the edges where {{{x=174/2=87}}} and at {{{x=-87}}}, {{{y=0}}}
Using the general vertex form,
{{{y=a(x-0)^2+27}}}
When {{{x=87}}},
{{{0=a(87)^2+27}}}
{{{a=-27/87^2}}}
{{{a=-27/7569}}}
So,
{{{y=-(27/7569)x^2+27}}}
When {{{x=12}}},
{{{y=-(27/7569)(12)^2+27}}}
{{{y=-3888/7569+27}}}
.
.
.
*[illustration fr1.JPG].
.
.
.