Question

Find the x-Intercept(s) and the coordinates of the vertex for the parabola y=x^2-2x-3. If there is more than one x-intercept, separate them with commas.

Answer 1

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:

`y=a(x-h)^2+k`

where the coordinates of the vertex is (h, k).

STEP 2: Execute

We will now resolve the polynomial and solve it.

`\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+((2)/(2))^2=3+((2)/(2))^2 \\ x^2-2x+1=3+1 \\ (x-1)^2=4 \end{gathered}`

Therefore, our vertex is at (1, -4)

Solving our equation,

`\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}`

The x intercepts are x = -1 and x = 3

These points are depicted in the graph as seen above.