Question

Find the quadratic function whose graph has a vertex at (3, 5) and passes through the point (1, -3). Express the function in both standard and general forms

Answer 1

Given vertex $(h,k)$ the equation of the parabola is:
$f(x)=a(x-h)^2+k$.
Our vertex is the point $(3,5)$  so the equation is:
$f(x)=a(x-3)^2+5$. It remains to find a. We will use the point (1,-3) like 
this:
Substitute x=1 and f(1)=-3 in the equation above we get the equation:
$-3=a(1-3)+5\\-3=-2a+5\\-2a=-8\\a=4$
Equation of the parabola then: $f(x)=4(x-3)^2+5$.
Expand in order to get the general form:
 $f(x)=4(x^2-3x+9)+5\\y=4x^2-12x+41$