Explanation
The vertex form of a quadratic function is:
[tex]\begin{gathered} y=a(x-h)^2+k \\ \text{ Where} \\ (h,k)\text{ is the vertex of the parabola } \end{gathered}[/tex]
The standard form of a quadratic function is:
[tex]y=ax^2+bx+c[/tex]
We can do the following steps to solve the exercise.
Step 1: We replace the values of h,k, x, and y into the vertex form of a quadratic equation, and we solve for a.
[tex]\begin{gathered} h=1 \\ k=-5 \\ x=3 \\ y=-1 \end{gathered}[/tex][tex]\begin{gathered} y=a(x-h)^{2}+k \\ -1=a(3-1)^2-5 \\ -1=a(2)^2-5 \\ -1=4a-5 \\ \text{ Add 5 from both sides} \\ -1+5=4a-5+5 \\ 4=4a \\ \text{ Divide by 4 from both sides} \\ \frac{4}{4}=\frac{4a}{4} \\ 1=a \end{gathered}[/tex]
Step 2: We replace the values of a,h, and k into the vertex form of a quadratic equation.
[tex]\begin{gathered} y=a(x-h)^{2}+k \\ y=1(x-1)^2-5 \end{gathered}[/tex]
Step 3: We expand the expression inside the parentheses and combine like terms to convert the function into its standard form.
[tex]\begin{gathered} y=1(x-1)^{2}-5 \\ y=(x-1)^2-5 \\ y=(x-1)(x-1)-5 \\ \text{ Apply the distributive property} \\ y=x(x-1)-1(x-1)-5 \\ y=x*x-x*1-1*x-1*-1-5 \\ y=x^2-x-x+1-5 \\ y=x^2-2x-4 \end{gathered}[/tex]Answer
The equation of the graph in standard form is:
[tex]y=x^{2}-2x-4[/tex]
