Question

Determine the zeros of the quadratic function y=x^2+15x+36 . (When listing the smaller zero, remember that negative values are always smaller than positive values.)(1 point) smaller zero: x=() ; greater zero: x=()

Answer 1

Determine the zeros of the quadratic function y=−x^2+16x−39 . (When listing the smaller zero, remember that negative values are always smaller than positive values.)(1 point) smaller zero: x=() ; greater zero: x=()

Answer 2

Determine the vertex and axis of symmetry of y=x^2−10x+3 .(1 point) vertex: (); axis of symmetry: x=()

Answer 3

what grade are you

Answer 4

What point is symmetrical to (−1,−3) for the graph of y=x^2+12x+8 ?(1 point) ()

Answer 5

What are the key points on the graph of y=x^2−16x+48 ? Name the vertex, x-intercepts, and y-intercept.(2 points) vertex: () (smaller) x-intercept: () (larger) x-intercept: () y-intercept: ()

Answer 6

To find the zeros of the quadratic function \( y = x^2 + 15x + 36 \), we can use the quadratic formula:

\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
\]

For the quadratic \( ax^2 + bx + c \), the coefficients are:
- \( a = 1 \)
- \( b = 15 \)
- \( c = 36 \)

Now, we can calculate the discriminant:

\[
b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot 36 = 225 - 144 = 81
\]

Since the discriminant is positive, we will have two real and distinct zeros.

Now substituting the values into the quadratic formula:

\[
x = \frac{{-15 \pm \sqrt{81}}}{2 \cdot 1}
\]

\[
x = \frac{{-15 \pm 9}}{2}
\]

Now we calculate the two possible values for \( x \):

1. \( x = \frac{{-15 + 9}}{2} = \frac{{-6}}{2} = -3 \)
2. \( x = \frac{{-15 - 9}}{2} = \frac{{-24}}{2} = -12 \)

Thus, the zeros of the function are \( x = -12 \) and \( x = -3 \).

Now we list them:

- Smaller zero: \( x = -12 \)
- Greater zero: \( x = -3 \)

So the final answer is:

- smaller zero: \( x = -12 \)
- greater zero: \( x = -3 \)

Answer 7

To find the zeros of the quadratic function \( y = -x^2 + 16x - 39 \), we will use the quadratic formula:

\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
\]

For the given quadratic \( ax^2 + bx + c \), the coefficients are:
- \( a = -1 \)
- \( b = 16 \)
- \( c = -39 \)

First, we calculate the discriminant:

\[
b^2 - 4ac = 16^2 - 4 \cdot (-1) \cdot (-39) = 256 - 156 = 100
\]

Since the discriminant is positive, there are two real and distinct zeros.

Now we substitute these values into the quadratic formula:

\[
x = \frac{{-16 \pm \sqrt{100}}}{2 \cdot (-1)}
\]

This simplifies to:

\[
x = \frac{{-16 \pm 10}}{-2}
\]

We will calculate the two possible values for \( x \):

1. For the plus sign:
\[
x = \frac{{-16 + 10}}{-2} = \frac{{-6}}{-2} = 3
\]

2. For the minus sign:
\[
x = \frac{{-16 - 10}}{-2} = \frac{{-26}}{-2} = 13
\]

Thus, the zeros of the function are \( x = 3 \) and \( x = 13 \).

Finally, we list them:

- Smaller zero: \( x = 3 \)
- Greater zero: \( x = 13 \)

So the final answer is:

- smaller zero: \( x = 3 \)
- greater zero: \( x = 13 \)

Answer 8

To find the vertex and axis of symmetry of the quadratic function \( y = x^2 - 10x + 3 \), we can use the vertex formula and the properties of quadratic functions.

### Vertex:
The vertex of a quadratic function in the form \( y = ax^2 + bx + c \) can be found using the formula:

\[
x = -\frac{b}{2a}
\]

For the given function, the coefficients are:
- \( a = 1 \)
- \( b = -10 \)

Substituting these values into the formula:

\[
x = -\frac{-10}{2 \cdot 1} = \frac{10}{2} = 5
\]

Now that we have the x-coordinate of the vertex, we can find the corresponding y-coordinate by substituting \( x = 5 \) back into the original function:

\[
y = (5)^2 - 10(5) + 3
\]
\[
y = 25 - 50 + 3 = -22
\]

Therefore, the vertex is:

\[
\text{vertex} = (5, -22)
\]

### Axis of Symmetry:
The axis of symmetry for a quadratic function is a vertical line that passes through the x-coordinate of the vertex. Thus, the axis of symmetry is given by:

\[
x = 5
\]

### Final Answers:
- vertex: \( (5, -22) \)
- axis of symmetry: \( x = 5 \)

Answer 9

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Answer 10

To find the point that is symmetrical to the point \((-1, -3)\) with respect to the graph of the quadratic function \(y = x^2 + 12x + 8\), we will follow these steps:

1. **Find the vertex of the parabola** since it acts as the axis of symmetry.
2. **Find the line of symmetry** which is a vertical line passing through the x-coordinate of the vertex.
3. **Calculate the symmetrical point.**

### Step 1: Calculate the Vertex
The vertex of a quadratic function in the form \(y = ax^2 + bx + c\) can be found using the formula:

\[
x = -\frac{b}{2a}
\]

For the function \(y = x^2 + 12x + 8\):
- \(a = 1\)
- \(b = 12\)

Plugging in these values:

\[
x = -\frac{12}{2 \cdot 1} = -\frac{12}{2} = -6
\]

Now, to find the corresponding \(y\)-coordinate of the vertex, substitute \(x = -6\) back into the equation:

\[
y = (-6)^2 + 12(-6) + 8
\]
\[
y = 36 - 72 + 8 = -28
\]

Thus, the vertex is at the point \((-6, -28)\).

### Step 2: Analyze the Symmetry
The axis of symmetry is the line \(x = -6\).

### Step 3: Find the Symmetrical Point
The x-coordinate of the original point is \(-1\). To find its symmetrical counterpart with respect to the line \(x = -6\), we calculate the distance from \(-1\) to \(-6\):

\[
\text{Distance} = -1 - (-6) = -1 + 6 = 5
\]

The symmetrical point will be the same distance from the line of symmetry, so we subtract this distance from \(-6\):

\[
\text{Symmetrical x-coordinate} = -6 - 5 = -11
\]

The y-coordinate remains the same because symmetry vertically concerning the x-axis does not change the y-value.

Thus, the symmetrical point is:

\[
\text{Symmetrical point} = (-11, -3)
\]

### Final Answer:
The point symmetrical to \((-1, -3)\) for the graph of \(y = x^2 + 12x + 8\) is \((-11, -3)\).

Answer 11

To find the key points on the graph of the quadratic function \( y = x^2 - 16x + 48 \), we need to determine the vertex, x-intercepts, and y-intercept. We will follow these steps:

### Step 1: Find the Vertex
The vertex of a quadratic function in the form \( y = ax^2 + bx + c \) can be found using the formula:

\[
x = -\frac{b}{2a}
\]

For the quadratic \( y = x^2 - 16x + 48 \):
- \( a = 1 \)
- \( b = -16 \)

Calculating the x-coordinate of the vertex:

\[
x = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8
\]

Next, substitute \( x = 8 \) back into the equation to find the y-coordinate of the vertex:

\[
y = (8)^2 - 16(8) + 48
\]
\[
y = 64 - 128 + 48 = -16
\]

So, the vertex is:

\[
\text{vertex} = (8, -16)
\]

### Step 2: Find the X-Intercepts
To find the x-intercepts, set \( y = 0 \):

\[
0 = x^2 - 16x + 48
\]

We can solve this quadratic equation using the quadratic formula:

\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
\]

Where \( a = 1 \), \( b = -16 \), and \( c = 48 \):

Calculating the discriminant:

\[
b^2 - 4ac = (-16)^2 - 4 \cdot 1 \cdot 48 = 256 - 192 = 64
\]

Since the discriminant is positive, there are two real and distinct x-intercepts.

Now plug in the values into the quadratic formula:

\[
x = \frac{{16 \pm \sqrt{64}}}{2 \cdot 1} = \frac{{16 \pm 8}}{2}
\]

Calculating the two possible values for \( x \):

1. For the plus sign:
\[
x = \frac{{16 + 8}}{2} = \frac{24}{2} = 12
\]

2. For the minus sign:
\[
x = \frac{{16 - 8}}{2} = \frac{8}{2} = 4
\]

Thus, the x-intercepts are \( x = 4 \) and \( x = 12 \).

### Step 3: Find the Y-Intercept
To find the y-intercept, set \( x = 0 \):

\[
y = (0)^2 - 16(0) + 48 = 48
\]

So, the y-intercept is:

\[
\text{y-intercept} = (0, 48)
\]

### Final Answers Summary:
- vertex: \( (8, -16) \)
- smaller x-intercept: \( (4, 0) \)
- larger x-intercept: \( (12, 0) \)
- y-intercept: \( (0, 48) \)

Thus, the answer is:
- vertex: \( (8, -16) \)
- (smaller) x-intercept: \( (4, 0) \)
- (larger) x-intercept: \( (12, 0) \)
- y-intercept: \( (0, 48) \)