Answer :
The vertex form of the parabola is
[tex]y=a(x-h)^2+k[/tex]Where (h, k) are the coordinates of its vertex
Since the vertex of the parabola is (5, -3), then
h = 5 and k = -3
Substitute them in the form above
[tex]\begin{gathered} y=a(x-5)^2+(-3) \\ y=a(x-5)^2-3 \end{gathered}[/tex]To find the value of (a) we will use the given point (1, 5)
Substitute x by 1 and y by 5
[tex]\begin{gathered} 5=a(1-5)^2-3 \\ 5=a(-4)^2-3 \\ 5=16a-3 \end{gathered}[/tex]Add both sides by 3
[tex]\begin{gathered} 5+3=16a-3+3 \\ 8=16a \end{gathered}[/tex]Divide both sides by 16 to find (a)
[tex]\begin{gathered} \frac{8}{16}=\frac{16a}{16} \\ \frac{1}{2}=a \\ a=\frac{1}{2} \end{gathered}[/tex]Substitute a by 1/2 in the equation above, then
a. The equation of the parabola is
[tex]y=\frac{1}{2}(x-5)^2-3[/tex]Let us substitute x and y in the equation by each answer
Since x = 0 and y = 3
[tex]\begin{gathered} 3=\frac{1}{2}(0-5)^2-3 \\ 3=12.5-3 \\ 3=9.5 \\ \text{LHS}\ne RHS \end{gathered}[/tex]Since x = -1 and y = 9
[tex]\begin{gathered} 9=\frac{1}{2}(-1-5)^2-3 \\ 9=\frac{1}{2}(-6)^2-3 \\ 9=\frac{1}{3}(36)-3 \\ 9=12-3 \\ 9=9 \\ \text{LHS}=\text{RHS} \end{gathered}[/tex]The point (-1, 9) lies on the parabola
The answer is b