Question 696155
{{{y=a(x-3/4)^2+k}}} is the equation of a parabola with axis of symmetry {{{x=3/4}}}
If we substitute the coordinates of point (0,1), we find
{{{1=a(0-3/4)^2+k}}}-->{{{1=9a/16+k}}}-->{{{1-k=9a/16}}}-->{{{16-16k=9a}}}-->{{{9a+16k=16}}}
If we substitute the coordinates of point (2,3), we find
{{{3=a(2-3/4)^2+k}}}-->{{{3=a(5/4)^2+k}}}-->{{{3=25a/16+k}}}-->{{{3-k=25a/16}}}-->{{{48-16k=25a}}}-->{{{25a+16k=48}}}
We now have the system
{{{system(25a+16k=48,9a+16k=16)}}}
Subtracting the second equation from the first we get
{{{25a+16k-(9a+16k)=48-16}}}-->{{{25a+16k-9a-16k=48-16}}}-->{{{16a=32}}}-->{{{highlight(a=2)}}}
Substituting that value back into {{{9a+16k=16}}} we get
{{{9*2+16k=16}}}-->{{{18+16k=16}}}-->{{{16k=16-18}}}-->{{{16k=-2}}}-->{{{highlight(k=-1/8)}}}
So the equation is
{{{highlight(y=2(x-3/4)^2-1/8)}}}
It can be transformed, like this:
{{{y=2(x^2-(3/2)x+9/16)-1/8}}}-->{{{y=2x^2-3x+9/8-1/8}}}-->{{{highlight(y=2x^2-3x+1)}}}