Question

Key points on the graph y = x^2 - 2x - 120

Answer 1

To analyze the quadratic function \( y = x^2 - 2x - 120 \), we can determine several key points, including the vertex, the x-intercepts, and the y-intercept.

### 1. **Vertex:**
The vertex of a parabola in the form \( y = ax^2 + bx + c \) can be found using the formula:
\[
x = -\frac{b}{2a}
\]
For the function \( y = x^2 - 2x - 120 \):
- \( a = 1 \)
- \( b = -2 \)

Calculating the x-coordinate of the vertex:
\[
x = -\frac{-2}{2 \cdot 1} = \frac{2}{2} = 1
\]

Now, substitute \( x = 1 \) back into the equation to find the y-coordinate:
\[
y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121
\]

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

### 2. **Y-Intercept:**
The y-intercept occurs when \( x = 0 \).
\[
y = (0)^2 - 2(0) - 120 = -120
\]
Thus, the y-intercept is:
\[
(0, -120)
\]

### 3. **X-Intercepts:**
To find the x-intercepts, set \( y = 0 \):
\[
0 = x^2 - 2x - 120
\]
We can solve this quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For this equation, \( a = 1 \), \( b = -2 \), and \( c = -120 \):
\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-120)}}{2 \cdot 1}
\]
Calculating the discriminant:
\[
(-2)^2 - 4 \cdot 1 \cdot (-120) = 4 + 480 = 484
\]
Now substitute back into the quadratic formula:
\[
x = \frac{2 \pm \sqrt{484}}{2} = \frac{2 \pm 22}{2}
\]
This gives us:
\[
x = \frac{24}{2} = 12 \quad \text{and} \quad x = \frac{-20}{2} = -10
\]
So the x-intercepts are:
\[
(12, 0) \quad \text{and} \quad (-10, 0)
\]

### Summary of Key Points:
1. **Vertex:** \((1, -121)\)
2. **Y-Intercept:** \((0, -120)\)
3. **X-Intercepts:** \((12, 0)\) and \((-10, 0)\)

These points are essential for sketching the graph of the quadratic function. The parabola opens upwards (since the coefficient of \( x^2 \) is positive), and the vertex indicates the minimum point of the graph.