Determine the zeros of the quadratic function y=x2+15x+36. (When listing the smaller zero, remember that negative values are always smaller than positive values.)(1 point)

Question

smaller zero: 𝑥=blank; greater zero: x=blank

Answer 1

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

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

In this function, the coefficients are:

• \( a = 1 \)
• \( b = 15 \)
• \( c = 36 \)

Now, let's calculate the discriminant (\( b^2 - 4ac \)): \[ b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot 36 = 225 - 144 = 81 \]

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

Now we can substitute the values into the quadratic formula: \[ x = \frac{-15 \pm \sqrt{81}}{2 \cdot 1} = \frac{-15 \pm 9}{2} \]

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

1. For the positive case: \[ x = \frac{-15 + 9}{2} = \frac{-6}{2} = -3 \]

2. For the negative case: \[ x = \frac{-15 - 9}{2} = \frac{-24}{2} = -12 \]

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

Now, we can list them:

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

Final answer:

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

Answer 2

Determine the zeros of the quadratic function y=−x2+16x−39
𝑦
=
−
𝑥
2
+
16
𝑥
−
39
. (When listing the smaller zero, remember that negative values are always smaller than positive values.)(1 point)
smaller zero: x=blak; greater zero: x=blank

Answer 3

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 this function, the coefficients are:

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

Let's first calculate the discriminant (\( b^2 - 4ac \)): \[ b^2 - 4ac = 16^2 - 4 \cdot (-1) \cdot (-39) = 256 - 156 = 100 \]

Since the discriminant is positive, there are two distinct real zeros. Now we can substitute the values into the quadratic formula:

\[ x = \frac{-16 \pm \sqrt{100}}{2 \cdot (-1)} = \frac{-16 \pm 10}{-2} \]

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

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

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

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

Now, we can list them:

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

Final answer:

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

Answer 4

Determine the vertex and axis of symmetry of y=x2−10x+3
𝑦
=
𝑥
2
−
10
𝑥
+
3
.(1 point)
vertex: (blank); axis of symmetry: x=blank

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 and the symmetry of the parabola.

1. Finding the Vertex

The \( x \)-coordinate of the vertex of a parabola given by the equation \( y = ax^2 + bx + c \) can be found using the formula:

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

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

Substituting these values in:

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

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

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

Thus, the vertex is:

\[ \text{Vertex: } (5, -22) \]

2. Finding the Axis of Symmetry

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

\[ x = 5 \]

Final Answers

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

Answer 6

What point is symmetrical to (−1,−3)
(
−
1
,
−
3
)
for the graph of y=x2+12x+8
𝑦
=
𝑥
2
+
12
𝑥
+
8
?(1 point)
(blank)

Answer 7

To find the symmetric point of \((-1, -3)\) with respect to the graph of the quadratic function \(y = x^2 + 12x + 8\), we will use the following steps:

1. Find the vertex of the parabola.
2. Calculate the distance from the point \((-1, -3)\) to the axis of symmetry.
3. Use this distance to find the symmetric point on the opposite side of the axis of symmetry.

Step 1: Find the Vertex

The vertex of the parabola given by \(y = ax^2 + bx + c\) can be found using:

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

where \(a = 1\) and \(b = 12\):

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

Now, find the \(y\)-coordinate by substituting \(x = -6\) back into the quadratic equation:

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

So, the vertex of the parabola is:

\[ (-6, -28) \]

Step 2: Calculate the Distance from the Point to the Axis of Symmetry

The axis of symmetry is the vertical line \(x = -6\). The distance from the point \((-1, -3)\) to this line is:

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

Step 3: Find the Symmetrical Point

The point that is symmetrical to \((-1, -3)\) will be located on the opposite side of the axis of symmetry. Since the distance to the axis of symmetry is 5, we subtract this distance from \(-6\):

\[ x_{\text{symmetric}} = -6 - 5 = -11 \]

Now we need to find the \(y\) value that corresponds to \(x = -11\):

\[ y = (-11)^2 + 12(-11) + 8 = 121 - 132 + 8 = -3 \]

Final Symmetrical Point

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

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

Answer 8

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

(smaller) x-intercept: (blank)

(larger) x-intercept: (blank)

y-intercept: (blank)

Answer 9

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

Step 1: Find the Vertex

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

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

For our function:

• \( a = 1 \)
• \( b = -16 \)

Calculating the x-coordinate of the vertex:

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

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

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

So, the vertex is:

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

Step 2: Find the x-intercepts

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

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

This can be solved using the quadratic formula:

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

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

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

Now, substituting into the quadratic formula:

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

Calculating the two values:

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

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

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

Step 3: Find the y-intercept

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

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

So the y-intercept is:

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

Final Answers

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