The quadratic equation:
[tex]y=x^2+6x-7[/tex]
has the form:
[tex]y=ax^2+bx+c[/tex]
with a = 1, b = 6, and c = -7.
We can find the x-intercepts with the help of the quadratic formula, as follows:
[tex]\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=\frac{-6+8}{2}=1 \\ x_2=\frac{-6-8}{2}=-7 \end{gathered}[/tex]
x-intercepts: 1, -7
The x-coordinate of the vertex can be found as follows:
[tex]\begin{gathered} x_v=-\frac{b}{2a} \\ x_v=-\frac{6}{2\cdot1} \\ x_v=-3 \end{gathered}[/tex]
The y-coordinate is found replacing the x-coordinate (Xv) into the equation of the parabola.
[tex]\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}[/tex]
Vertex: (-3, -16)
