Answer:
The zeros of the function are
[tex]x=2+2\sqrt{5}[/tex]
[tex]x=2-2\sqrt{5}[/tex]
The graph in the attached figure
Step-by-step explanation:
we have
[tex]y=x^{2}-4x-16[/tex]
This is a vertical parabola open upward (the leading coefficient is positive)
The vertex is a minimum
Remember that
The zeros of the function are the values of x when the value of y is equal to zero
For y=0
[tex]x^{2}-4x-16=0[/tex]
Move the constant term to the right side
[tex]x^{2}-4x=16[/tex]
Complete the square
[tex]x^{2}-4x+2^2=16+2^2[/tex]
[tex]x^{2}-4x+4=20[/tex]
Rewrite as perfect squares
[tex](x-2)^{2}=20[/tex] ---> the vertex is the point (2,-20)
take square root both sides
[tex]x-2=\pm\sqrt{20}[/tex]
[tex]x=2\pm\sqrt{20}[/tex]
Simplify
[tex]x=2\pm2\sqrt{5}[/tex]
The zeros of the function are
[tex]x=2+2\sqrt{5}[/tex]
[tex]x=2-2\sqrt{5}[/tex]
using a graphing tool
The graph in the attached figure
