The vertex of the considered function is found to be on (1, -9)
What is vertex form of a quadratic equation?
If a quadratic equation is written in the form
[tex]y=a(x-h)^2 + k[/tex]
then it is called to be in vertex form. It is called so because when you plot this equation's graph, you will see vertex point(peak point) is on [tex](h,k)[/tex]
We first take out coefficient of x squared, and then inside the bracket, we try to make perfect square like situation.
For this case, the function given is:
[tex]y = x^2 - 2x - 8[/tex]
The coefficient of [tex]x^2[/tex] is 1, so no need to take it out.
Converting it to the vertex form, we get:
[tex]y = x^2 - 2x - 8\\y = x^2 - 2\times 1 \times x + 1^2 - 1^2 -8[/tex]
(we did so as to use [tex]a^2 + 2ab + b^2 = (a+b)^2[/tex] formula)
[tex]y = x^2 - 2\times 1 \times x + 1^2 - 1^2 -8\\y = (x-1)^2 - 1 - 8\\y = (x-1)^2 -9[/tex]
Comparing this with the vertex form [tex]y=a(x-h)^2 + k[/tex], we get;
h = 1. k = -9
Thus, the vertex of the considered function is on the point (1,-9)
Learn more about vertex form of a quadratic equation here:
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