Use the image to answer the question. Identify the vertex on the graph. (1 point) Responses (5,0) left parenthesis 5 comma 0 right parenthesis (0,−5) left parenthesis 0 comma negative 5 right parenthesis (−5,0) left parenthesis negative 5 comma 0 right parenthesis x=−5 x equals negative 5

Without seeing the image, I'm unable to identify the vertex from the graph directly. However, if you provide the information or context about the graph, such as its equation or description, I can help identify the vertex. In typical scenarios, the vertex of a parabola in the form \( y = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \).

Determine the zero(s) of the quadratic function y=−5x2+16x.(1 point)

Responses

(0,5)(0,16)
left parenthesis 0 comma 5 right parenthesis left parenthesis 0 comma 16 right parenthesis

There are no zeros.
There are no zeros.

(−3.2,0)
left parenthesis negative 3.2 comma 0 right parenthesis

(0,0)(3.2,0)
left parenthesis 0 comma 0 right parenthesis left parenthesis 3.2 comma 0 right parenthesis

To determine the zeros of the quadratic function \( y = -5x^2 + 16x \), we need to set the equation equal to zero and solve for \( x \):

\[
-5x^2 + 16x = 0
\]

Factor out \( x \):

\[
x(-5x + 16) = 0
\]

This gives us two factors to set equal to zero:

1. \( x = 0 \)
2. \( -5x + 16 = 0 \)

Now, solve the second equation:

\[
-5x + 16 = 0 \implies 5x = 16 \implies x = \frac{16}{5} = 3.2
\]

Thus, the zeros of the function are \( x = 0 \) and \( x = 3.2 \). The corresponding points on the graph are \( (0, 0) \) and \( (3.2, 0) \).

**The correct response is: (0,0) (3.2,0).**

Identify the vertex of y=4(x−25)2−61.(1 point)

Responses

(−25,−61)
left parenthesis negative 25 comma negative 61 right parenthesis

(−100,−61)
left parenthesis negative 100 comma negative 61 right parenthesis

(4,−25)
left parenthesis 4 comma negative 25 right parenthesis

(25,−61)
left parenthesis 25 comma negative 61 right parenthesis

The given equation is in vertex form:

\[
y = a(x - h)^2 + k
\]

where \((h, k)\) is the vertex. In this case, the equation is:

\[
y = 4(x - 25)^2 - 61
\]

From this, we can see:

- \(h = 25\)
- \(k = -61\)

Thus, the vertex is \((25, -61)\).

**The correct response is: (25, -61).**

What are the key points on the graph of y=x2−2x−120? Name the vertex, x-intercept(s), and y-intercept.(1 point)

Responses

x-intercept: (1,−121)

y-intercept: (0,−120)

vertex: (−10,0) (12,0)

x -intercept: left parenthesis 1 comma negative 121 right parenthesis y -intercept: left parenthesis 0 comma negative 120 right parenthesis vertex: left parenthesis negative 10 comma 0 right parenthesis left parenthesis 12 comma 0 right parenthesis

x-intercepts: (−10,0) (12,0)

y-intercept: (0,−120)

vertex: (1,−121)

x -intercepts: left parenthesis negative 10 comma 0 right parenthesis left parenthesis 12 comma 0 right parenthesis y -intercept: left parenthesis 0 comma negative 120 right parenthesis vertex: left parenthesis 1 comma negative 121 right parenthesis

x-intercepts: (−10,0) (12,0)

y-intercept: (1,−121)

vertex: (0,−120)

x -intercepts: left parenthesis negative 10 comma 0 right parenthesis left parenthesis 12 comma 0 right parenthesis y-intercept: left parenthesis 1 comma negative 121 right parenthesis vertex: left parenthesis 0 comma negative 120 right parenthesis

x-intercept: (0,−120)

y-intercept: (−10,0) (12,0)

vertex: (1,−121)

To find the key points on the graph of the quadratic function \( y = x^2 - 2x - 120 \), we can:

1. **Find the vertex** using the formula \( x = -\frac{b}{2a} \), where \( a = 1 \) and \( b = -2 \) in the general form \( y = ax^2 + bx + c \).

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

Now, substitute \( x = 1 \) back into the equation to find \( y \):

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

So the vertex is \( (1, -121) \).

2. **Find the x-intercepts** by setting \( y = 0 \):

\[
0 = x^2 - 2x - 120
\]

Factoring the equation:

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

Setting each factor to zero gives:

\[
x - 12 = 0 \implies x = 12 \quad \text{and} \quad x + 10 = 0 \implies x = -10
\]

Thus, the x-intercepts are \( (-10, 0) \) and \( (12, 0) \).

3. **Find the y-intercept** by setting \( x = 0 \):

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

Thus, the y-intercept is \( (0, -120) \).

### Summary of Key Points

- **Vertex**: \( (1, -121) \)
- **X-intercepts**: \( (-10, 0) \) and \( (12, 0) \)
- **Y-intercept**: \( (0, -120) \)

**The correct response is:**

- x-intercepts: \( (-10, 0) \) and \( (12, 0) \)
- y-intercept: \( (0, -120) \)
- vertex: \( (1, -121) \)