Question

What are the key points on the graph of y=x2−2x−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 find the key points on the graph of the quadratic function \(y = x^2 - 2x - 120\), we'll follow these steps:

1. **Find the x-intercepts**: To find the x-intercepts, set \(y = 0\) and solve the equation \(x^2 - 2x - 120 = 0\).

The equation can be factored as follows:

\[
x^2 - 2x - 120 = (x - 12)(x + 10) = 0
\]

Setting each factor to zero gives:

\[
x - 12 = 0 \implies x = 12
\]
\[
x + 10 = 0 \implies x = -10
\]

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

2. **Find the y-intercept**: To find the y-intercept, set \(x = 0\) in the original equation:

\[
y = 0^2 - 2(0) - 120 = -120
\]

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

3. **Find the vertex**: The vertex of a parabola in standard form \(y = ax^2 + bx + c\) can be found using the formula for the x-coordinate of the vertex:

\[
x = -\frac{b}{2a}
\]
Here, \(a = 1\) and \(b = -2\):

\[
x = -\frac{-2}{2(1)} = \frac{2}{2} = 1
\]

Now substitute \(x = 1\) back into the original equation to find the y-coordinate of the vertex:

\[
y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121
\]

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

In summary, the key points are:
- **x-intercepts**: \((-10, 0)\) and \((12, 0)\)
- **y-intercept**: \((0, -120)\)
- **vertex**: \((1, -121)\)