STEP 1: Identify and Set-Up
We are given the equation of the parabola in the polynomial. However, there exists a vertex form through which we can obtain the coordinates of our vertex. The solution of the quadratic equation gives us the x intercepts.
The general equation of a parabola in vertex form is:
[tex]y=a(x-h)^2+k[/tex]where the coordinates of the vertex is (h, k).
STEP 2: Execute
We will now resolve the polynomial and solve it.
[tex]\begin{gathered} y=x^2-2x-3_{} \\ We\text{ equate the function to zero first} \\ x^2-2x-3=0 \\ \text{Adding 3 to both sides and completing the square gives us} \\ x^2-2x+(\frac{2}{2})^2=3+(\frac{2}{2})^2 \\ x^2-2x+1=3+1 \\ (x-1)^2=4 \end{gathered}[/tex]Therefore, our vertex is at (1, -4)
Solving our equation,
[tex]\begin{gathered} (x-1)^2=4 \\ \text{Getting the square root of both sides gives:} \\ (x-1)=\sqrt[]{4}=2 \\ x-1=\pm2 \\ x=1\pm2 \\ x=3\text{ or -1} \end{gathered}[/tex]The x intercepts are x = -1 and x = 3
These points are depicted in the graph as seen above.
