We have to find the equation of the quadratic function.
We know the vertex, located in (4,6), and one point (-5,-2).
The x-coordinate of the vertex (4) is equal to -b/2a, being a the quadratic coefficient and b the linear coefficient.
Now, we have 2 points to define the 3 parameters, so one of the parameters is undefined.
[tex]y=ax^2+bx+c[/tex]We start with the vertex, that we know that is:
[tex]\begin{gathered} x=-\frac{b}{2a}=4 \\ -b=4\cdot2a=8a \\ b=-8a \end{gathered}[/tex]Then, we can write the equation as:
[tex]y=ax^2-8ax+c=a(x^2-8x)+c[/tex]If we replace the point (-5,-2) in the equation, we get:
[tex]\begin{gathered} -2=a((-5)^2-8\cdot(-5))+c \\ -2=a(25+40)+c \\ -2=65a+c \\ c=-2-65a \end{gathered}[/tex]We replace the vertex coordinates and get:
[tex]\begin{gathered} 6=a(4^2-8\cdot4)+c \\ 6=a(16-32)+(-2-65a) \\ 6=-16a-2-65a \\ 6=-81a-2 \\ 81a=-2-6 \\ a=-\frac{8}{81}\approx-0.01 \end{gathered}[/tex]Then, the linear coefficient b is:
[tex]b=-8a=-8\cdot(-\frac{8}{81})=\frac{64}{81}\approx0.79[/tex]And the constant term is:
[tex]c=-2-65a=-2-65\cdot(-\frac{8}{81})=-2+\frac{520}{81}=\frac{-162+520}{81}=\frac{358}{81}\approx4.42[/tex]The quadratic coefficient is a=-0.01
The linear coefficient is b=0.79
the constant term is c=4.42
