Determine the zeros of the quadratic function y=−x^2+16x−39

Question

. (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

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

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

Here, \( a = -1 \), \( b = 16 \), and \( c = -39 \).

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

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

Now substituting the values into the quadratic formula:

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

Calculating the square root and the formula:

\[ \sqrt{100} = 10 \] \[ x = \frac{-16 \pm 10}{-2} \]

This gives us two solutions:

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

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

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

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

Therefore the zeros of the quadratic function are \( x = 3 \) and \( x = 13 \).

To summarize:

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

Final answer:

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

Answer 2

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

Answer 3

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:

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

Here, \( a = 1 \) and \( b = -10 \).

1. Finding the x-coordinate of the vertex:

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

2. Finding the y-coordinate of the vertex:

To find the y-coordinate, substitute \( x = 5 \) back into the function:

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

Thus, the vertex is at the point \( (5, -22) \).

3. Finding the axis of symmetry:

The axis of symmetry is given by the x-coordinate of the vertex:

\[ x = 5 \]

To summarize:

• Vertex: \( (5, -22) \)
• Axis of symmetry: \( x = 5 \)

Final answer:

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

Answer 4

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

Answer 5

To find the point symmetrical to \((-1, -3)\) for the graph of the quadratic function \(y = x^2 + 12x + 8\), we first need to determine the vertex (which acts as the axis of symmetry) of the parabola represented by the quadratic function.

Step 1: Find the vertex of \(y = x^2 + 12x + 8\)

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

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

In this case:

• \(a = 1\)
• \(b = 12\)

Calculating the x-coordinate of the vertex:

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

Step 2: Find the y-coordinate of the vertex

We can substitute \(x = -6\) back into the function to find the y-coordinate:

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

So, the vertex of the parabola is at \((-6, -28)\).

Step 3: Find the symmetrical point

The axis of symmetry for the parabola is the vertical line \(x = -6\). To find the symmetrical point of \((-1, -3)\) with respect to this line, we can calculate its reflection.

1. Calculate the distance from the point to the axis of symmetry:
   • The distance from \(-1\) to \(-6\) is \( |-1 - (-6)| = |-1 + 6| = |5| = 5 \).
2. Reflect that distance to find the symmetrical point:
   • To find the x-coordinate of the symmetrical point, subtract this distance from the x-coordinate of the axis of symmetry:
   • The x-coordinate of the symmetrical point is \(-6 - 5 = -11\).
3. The y-coordinate remains the same:
   • The y-coordinate of the symmetrical point is still \(-3\) since it only reflects horizontally.

Final answer

The point symmetrical to \((-1, -3)\) with respect to the graph of \(y = x^2 + 12x + 8\) is:

\[ (-11, -3) \]

So, the final answer is:

\[ (-11, -3) \]

Answer 6

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 7

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

Step 1: Find the vertex

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

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

Here, \( a = 1 \) and \( b = -16 \).

Calculating the x-coordinate of the vertex:

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

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

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

Thus, the vertex is at the point:

\[ \text{Vertex: } (8, -16) \]

Step 2: Find the x-intercepts

To find the x-intercepts, we set \( y = 0 \) and solve for \( x \):

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

Using the quadratic formula:

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

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

Calculating the discriminant \( b^2 - 4ac \):

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

Now substituting into the quadratic formula:

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

Calculating the square root:

\[ \sqrt{64} = 8 \]

Calculating the two x-intercepts:

1. For \( x = \frac{16 + 8}{2} = \frac{24}{2} = 12 \)
2. For \( x = \frac{16 - 8}{2} = \frac{8}{2} = 4 \)

Thus, the x-intercepts are:

\[ \text{Smaller x-intercept: } (4, 0) \] \[ \text{Larger x-intercept: } (12, 0) \]

Step 3: Find the y-intercept

To find the y-intercept, we set \( x = 0 \):

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

Thus, the y-intercept is:

\[ \text{Y-intercept: } (0, 48) \]

Summary of results

• Vertex: \( (8, -16) \)
• Smaller x-intercept: \( (4, 0) \)
• Larger x-intercept: \( (12, 0) \)
• Y-intercept: \( (0, 48) \)

Final answer format:

• vertex: \( (8, -16) \)
• (smaller) x-intercept: \( (4, 0) \)
• (larger) x-intercept: \( (12, 0) \)
• y-intercept: \( (0, 48) \)