Question

Find the x-intercept(s) and the coordinates of the vertex for the parabola y=x^2+6x-7. If there is more than one x-intercept, separate them with commas.

Answer 1

The quadratic equation:

`y=x^2+6x-7`

has the form:

`y=ax^2+bx+c`

with a = 1, b = 6, and c = -7.

We can find the x-intercepts with the help of the quadratic formula, as follows:

`\begin{gathered} x_(1,2)=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x_(1,2)=\frac{-6\pm\sqrt[]{6^2-4\cdot1\cdot(-7)}}{2\cdot1} \\ x_(1,2)=\frac{-6\pm\sqrt[]{64}}{2} \\ x_1=(-6+8)/(2)=1 \\ x_2=(-6-8)/(2)=-7 \end{gathered}`

x-intercepts: 1, -7

The x-coordinate of the vertex can be found as follows:

`\begin{gathered} x_v=-(b)/(2a) \\ x_v=-(6)/(2\cdot1) \\ x_v=-3 \end{gathered}`

The y-coordinate is found replacing the x-coordinate (Xv) into the equation of the parabola.

`\begin{gathered} y_v=x^2_v+6x_v-7 \\ y_v=(-3)^2+6\cdot(-3)_{}-7 \\ y_v=9-18-7 \\ y_v=-16 \end{gathered}`

Vertex: (-3, -16)