Question

11. What are the key points on the graph of y=x2−2x−120

=

2
−
2

−
120
? Name the vertex, x-intercepts, and y-intercept. Write your answers as ordered pairs. For the x-intercepts, write the leftmost point first. (4 points)
x-intercepts: and

y-intercept:

vertex:

Answer 1

To determine the key points of the quadratic function \(y = x^2 - 2x - 120\), we can follow these steps:

1. **Vertex**:
The vertex of a quadratic function \(y = ax^2 + bx + c\) can be found using the formula for the x-coordinate of the vertex:
\[
x = -\frac{b}{2a}
\]
Here, \(a = 1\) and \(b = -2\).

\[
x = -\frac{-2}{2 \times 1} = \frac{2}{2} = 1
\]

To find the y-coordinate of the vertex, we substitute \(x = 1\) back into the equation:
\[
y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121
\]

Thus, the vertex is:
\[
\text{vertex: } (1, -121)
\]

2. **Y-intercept**:
The y-intercept can be found by evaluating \(y\) when \(x = 0\):
\[
y = (0)^2 - 2(0) - 120 = -120
\]

Therefore, the y-intercept is:
\[
\text{y-intercept: } (0, -120)
\]

3. **X-intercepts**:
The x-intercepts occur when \(y = 0\):
\[
0 = x^2 - 2x - 120
\]
We can factor this quadratic equation or use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Here, \( a = 1 \), \( b = -2 \), and \( c = -120 \).

The discriminant \(D\) is:
\[
D = b^2 - 4ac = (-2)^2 - 4(1)(-120) = 4 + 480 = 484
\]

Now applying the quadratic formula:
\[
x = \frac{-(-2) \pm \sqrt{484}}{2(1)} = \frac{2 \pm 22}{2}
\]
This gives us two solutions:
\[
x_1 = \frac{24}{2} = 12 \quad \text{(rightmost intercept)}
\]
\[
x_2 = \frac{-20}{2} = -10 \quad \text{(leftmost intercept)}
\]

Therefore, the x-intercepts are:
\[
\text{x-intercepts: } (-10, 0) \text{ and } (12, 0)
\]

**Summary of key points**:
- x-intercepts: \((-10, 0)\) and \((12, 0)\)
- y-intercept: \((0, -120)\)
- vertex: \((1, -121)\)

Please note to use the x-intercepts in the format requested by providing the leftmost point first.