Question 921569
The Vertex of This Parabola is at (3,5). When the y-value is 6, the x-value is
-1 what is the coefficient of the squared term in the parabola's equation?
<pre>
{{{y = a(x-h)^2+k}}}

The Vertex of This Parabola is at (3,5)

So (h,k) = (3,5)

{{{y = a(x-3)^2+5}}}

When the y-value is 6, the x-value is -1

{{{6 = a(-1-3)^2+5}}}
{{{6 = a(-4)^2+5}}}
{{{6 = a(16)+5}}}
{{{6 = 16a+5}}}
{{{1 = 16a}}}
{{{1/16 = a}}}  <-- that's the answer as we shall see:

{{{y = expr(1/16)(x-3)^2+5}}}
{{{y = expr(1/16)(x^2-6x+9)+5}}}
{{{y = expr(1/16)x^2-expr(3/8)x+9/16+5}}}
{{{y = expr(1/16)x^2-expr(3/8)x+9/16+80/16}}}
{{{y = expr(1/16)x^2-expr(3/8)x+89/16}}}

So the coefficient of the squared term is {{{1/16}}}.

Edwin</pre>