Question 362938
Start with the vertex form, {{{y=a(x-h)^2+k}}} where (h,k) is the vertex.
{{{y=a(x-4)^2-5}}}
Now use the point (0,4) to solve for {{{a}}}.
{{{4=a(0-4)^2-5}}}
{{{16a=9}}}
{{{a=9/16}}}
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{{{y=(9/16)(x-4)^2-5}}}
{{{y=(9/16)(x^2-8x+16)-5}}}
{{{y=(9/16)x^2-(9/8)x+9-5}}}
{{{highlight(y=(9/16)x^2-(9/8)x+4)}}}