Question

Determine the zeros of the quadratic function y=−x2+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=
Remaining Attempts : 3

Answer 1

Determine the vertex and axis of symmetry of y=x2−10x+3
.(1 point)
vertex: (
); axis of symmetry: x=

Answer 2

What point is symmetrical to (−1,−3)
for the graph of y=x2+12x+8
?(1 point)
(
)

Answer 3

What are the key points on the graph of y=x2−16x+48
? Name the vertex, x-intercepts, and y-intercept.(2 points)
vertex: (
)

(smaller) x-intercept: (
)

(larger) x-intercept: (
)

y-intercept: (
)

Remaining Attempts : 3
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Answer 4

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

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

In the given quadratic \( y = -x^2 + 16x - 39 \), the coefficients are:

- \( a = -1 \)
- \( b = 16 \)
- \( c = -39 \)

First, we calculate the discriminant \( b^2 - 4ac \):

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

Now, substituting into the quadratic formula:

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

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

Now calculating the two possible values for \( x \):

1. For \( x = \frac{-16 + 10}{-2} \):

\[
x = \frac{-6}{-2} = 3
\]

2. For \( x = \frac{-16 - 10}{-2} \):

\[
x = \frac{-26}{-2} = 13
\]

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

When listing the smaller zero first:

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

So the final answers are:

**smaller zero: x = 3; greater zero: x = 13**

Answer 5

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

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

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

In this case, the coefficients are:

- \( a = 1 \)
- \( b = -10 \)
- \( c = 3 \)

Now, we can calculate the x-coordinate of the vertex:

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

Next, we substitute \( x = 5 \) back into the original quadratic equation to find the y-coordinate of the vertex:

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

Thus, the vertex is:

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

The axis of symmetry is a vertical line that passes through the vertex, given by the equation:

\[
x = 5
\]

So, the final answers are:

- **vertex: (5, -22)**
- **axis of symmetry: x = 5**

Answer 6

To find the point that is symmetrical to \((-1, -3)\) for the graph of the quadratic function \(y = x^2 + 12x + 8\), we first need to determine the vertex of the parabola, as the axis of symmetry will pass through this point.

### 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 for the x-coordinate of the vertex:

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

For the function \(y = x^2 + 12x + 8\):

- \(a = 1\)
- \(b = 12\)
- \(c = 8\)

Using the vertex formula:

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

Now, we calculate the y-coordinate of the vertex by substituting \(x = -6\) back into the function:

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

Thus, the vertex is at:

\[
\text{Vertex} = (-6, -28)
\]

### Step 2: Find the point symmetrical to \((-1, -3)\)

The axis of symmetry for the parabola is a vertical line given by the equation \(x = -6\). We need to find the point that is symmetrical to \((-1, -3)\) with respect to this axis.

The distance from the point \((-1, -3)\) to the axis of symmetry \(x = -6\) can be calculated as follows:

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

To find the symmetrical point, we move 5 units to the left of the axis of symmetry \(x = -6\):

\[
-6 - 5 = -11
\]

The y-coordinate remains the same, so the symmetrical point will be \((-11, -3)\).

### Final Answer

Thus, the point that is symmetrical to \((-1, -3)\) is:

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

Answer 7

To find the key points of the quadratic function \(y = x^2 - 16x + 48\), we will determine the vertex, x-intercepts, and y-intercept.

### Step 1: Find the Vertex

The vertex can be found using the formula for the x-coordinate of the vertex:

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

For the given equation:

- \(a = 1\)
- \(b = -16\)

Calculating the x-coordinate:

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

Now substituting \(x = 8\) back into the function to find the y-coordinate:

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

Thus, the vertex is:

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

### Step 2: Find the x-intercepts

The x-intercepts are found by setting \(y = 0\):

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

We can factor the quadratic:

\[
0 = (x - 6)(x - 8)
\]

Thus, the x-intercepts are:

\[
x = 6 \quad \text{and} \quad x = 8
\]

- The smaller x-intercept is at \((6, 0)\).
- The larger x-intercept is at \((8, 0)\).

### Step 3: Find the y-intercept

The y-intercept is found by setting \(x = 0\):

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

Thus, the y-intercept is at:

\[
(0, 48)
\]

### Summary of Key Points

- **Vertex:** \((8, -16)\)
- **Smaller x-intercept:** \((6, 0)\)
- **Larger x-intercept:** \((8, 0)\)
- **y-intercept:** \((0, 48)\)

### Final Answers

- **vertex:** \((8, -16)\)
- **(smaller) x-intercept:** \((6, 0)\)
- **(larger) x-intercept:** \((8, 0)\)
- **y-intercept:** \((0, 48)\)