To find the [tex]\(x\)[/tex]-intercepts and the coordinates of the vertex for the quadratic equation [tex]\(y = x^2 + 6x - 7\)[/tex], let's go through the process step-by-step.
### Finding the [tex]\(x\)[/tex]-Intercepts
The [tex]\(x\)[/tex]-intercepts are the points where the parabola intersects the [tex]\(x\)[/tex]-axis. These occur where [tex]\(y = 0\)[/tex].
1. Set [tex]\(y = 0\)[/tex]:
[tex]\[0 = x^2 + 6x - 7\][/tex]
2. Solve the quadratic equation [tex]\(x^2 + 6x - 7 = 0\)[/tex].
To find the roots, we can use the quadratic formula:
[tex]\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\][/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(c = -7\)[/tex]:
[tex]\[
x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{-6 \pm \sqrt{36 + 28}}{2}
\][/tex]
[tex]\[
x = \frac{-6 \pm \sqrt{64}}{2}
\][/tex]
[tex]\[
x = \frac{-6 \pm 8}{2}
\][/tex]
This gives us two solutions:
[tex]\[
x = \frac{-6 + 8}{2} = \frac{2}{2} = 1
\][/tex]
[tex]\[
x = \frac{-6 - 8}{2} = \frac{-14}{2} = -7
\][/tex]
So, the [tex]\(x\)[/tex]-intercepts are [tex]\(x = 1\)[/tex] and [tex]\(x = -7\)[/tex].
### Finding the Vertex
The vertex of a parabola given by the equation [tex]\(y = ax^2 + bx + c\)[/tex] can be found using the formula for the [tex]\(x\)[/tex]-coordinate of the vertex:
[tex]\[
x = -\frac{b}{2a}
\][/tex]
Here, [tex]\(a = 1\)[/tex] and [tex]\(b = 6\)[/tex]:
[tex]\[
x = -\frac{6}{2 \cdot 1} = -\frac{6}{2} = -3
\][/tex]
To find the [tex]\(y\)[/tex]-coordinate of the vertex, substitute [tex]\(x = -3\)[/tex] into the original equation [tex]\(y = x^2 + 6x - 7\)[/tex]:
[tex]\[
y = (-3)^2 + 6(-3) - 7
\][/tex]
[tex]\[
y = 9 - 18 - 7
\][/tex]
[tex]\[
y = -16
\][/tex]
So, the coordinates of the vertex are [tex]\((-3, -16)\)[/tex].
### Summary
Hence, the [tex]\(x\)[/tex]-intercepts and the coordinates of the vertex are:
[tex]\(x\)[/tex]-intercept(s): [tex]\(-7, 1\)[/tex]
Vertex: [tex]\((-3, -16)\)[/tex]
