Question
What are the key points on the graph of $y=x^{2}-2x-120$ ? Name the vertex x-intercept(s),
and y-intercept. (1 point)
x-intercept: $(1,-121)$
y-intercept: $(0,-120)$
vertex: $(-10,0)(12,0)$
x-intercepts: $(-10,0)(12,0)$
y-intercept: $(1,-121)$
vertex: $(0,-120)$
x-intercepts: $(-10,0)(12,0)$
y-intercept: $(0,-120)$
vertex: $(1,-121)$
x-intercept: $(0,-120)$
y-intercept: $(-10,0)(12,0)$
vertex: $(1,-121)$
and y-intercept. (1 point)
x-intercept: $(1,-121)$
y-intercept: $(0,-120)$
vertex: $(-10,0)(12,0)$
x-intercepts: $(-10,0)(12,0)$
y-intercept: $(1,-121)$
vertex: $(0,-120)$
x-intercepts: $(-10,0)(12,0)$
y-intercept: $(0,-120)$
vertex: $(1,-121)$
x-intercept: $(0,-120)$
y-intercept: $(-10,0)(12,0)$
vertex: $(1,-121)$
Answers
GPT-4o mini
To analyze the quadratic function \( y = x^2 - 2x - 120 \), we can find the vertex, x-intercepts, and y-intercept step by step.
1. **Finding the Vertex:**
The vertex of a parabola given by the equation \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex:
\[
x = -\frac{b}{2a}
\]
Here, \( a = 1 \) and \( b = -2 \). Plugging in these values:
\[
x = -\frac{-2}{2 \cdot 1} = \frac{2}{2} = 1
\]
Now, substituting \( x = 1 \) back into the equation to find the y-coordinate:
\[
y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121
\]
Therefore, the vertex is at \( (1, -121) \).
2. **Finding the x-intercepts:**
To find the x-intercepts, we set \( y = 0 \):
\[
0 = x^2 - 2x - 120
\]
This can be factored into:
\[
0 = (x - 12)(x + 10)
\]
Therefore, the x-intercepts are:
\[
x - 12 = 0 \quad \Rightarrow \quad x = 12 \quad \text{and} \quad x + 10 = 0 \quad \Rightarrow \quad x = -10
\]
Thus, the x-intercepts are \( (-10, 0) \) and \( (12, 0) \).
3. **Finding the y-intercept:**
The y-intercept occurs where \( x = 0 \):
\[
y = (0)^2 - 2(0) - 120 = -120
\]
Therefore, the y-intercept is \( (0, -120) \).
4. **Summary of Key Points:**
- Vertex: \( (1, -121) \)
- x-intercepts: \( (-10, 0) \) and \( (12, 0) \)
- y-intercept: \( (0, -120) \)
Among the choices provided, the correct set of key points for the graph of the function \( y = x^2 - 2x - 120 \) is:
- Vertex: \( (1, -121) \)
- x-intercepts: \( (-10, 0) \) and \( (12, 0) \)
- y-intercept: \( (0, -120) \)
Thus, the correct answer is:
- Vertex: \( (1, -121) \)
- x-intercepts: \( (-10, 0) \) and \( (12, 0) \)
- y-intercept: \( (0, -120) \)
1. **Finding the Vertex:**
The vertex of a parabola given by the equation \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex:
\[
x = -\frac{b}{2a}
\]
Here, \( a = 1 \) and \( b = -2 \). Plugging in these values:
\[
x = -\frac{-2}{2 \cdot 1} = \frac{2}{2} = 1
\]
Now, substituting \( x = 1 \) back into the equation to find the y-coordinate:
\[
y = (1)^2 - 2(1) - 120 = 1 - 2 - 120 = -121
\]
Therefore, the vertex is at \( (1, -121) \).
2. **Finding the x-intercepts:**
To find the x-intercepts, we set \( y = 0 \):
\[
0 = x^2 - 2x - 120
\]
This can be factored into:
\[
0 = (x - 12)(x + 10)
\]
Therefore, the x-intercepts are:
\[
x - 12 = 0 \quad \Rightarrow \quad x = 12 \quad \text{and} \quad x + 10 = 0 \quad \Rightarrow \quad x = -10
\]
Thus, the x-intercepts are \( (-10, 0) \) and \( (12, 0) \).
3. **Finding the y-intercept:**
The y-intercept occurs where \( x = 0 \):
\[
y = (0)^2 - 2(0) - 120 = -120
\]
Therefore, the y-intercept is \( (0, -120) \).
4. **Summary of Key Points:**
- Vertex: \( (1, -121) \)
- x-intercepts: \( (-10, 0) \) and \( (12, 0) \)
- y-intercept: \( (0, -120) \)
Among the choices provided, the correct set of key points for the graph of the function \( y = x^2 - 2x - 120 \) is:
- Vertex: \( (1, -121) \)
- x-intercepts: \( (-10, 0) \) and \( (12, 0) \)
- y-intercept: \( (0, -120) \)
Thus, the correct answer is:
- Vertex: \( (1, -121) \)
- x-intercepts: \( (-10, 0) \) and \( (12, 0) \)
- y-intercept: \( (0, -120) \)