etermine the zeros of the quadratic function y=x2+15x+36

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 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 case, the coefficients are:

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

Now, we can plug these values into the formula:

1. Calculate the discriminant: \[ b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot 36 = 225 - 144 = 81 \]

2. Use the quadratic formula: \[ x = \frac{-15 \pm \sqrt{81}}{2 \cdot 1} = \frac{-15 \pm 9}{2} \]

3. Calculate the two possible values for \( x \):

   • For the positive case: \[ x = \frac{-15 + 9}{2} = \frac{-6}{2} = -3 \]
   • For the negative case: \[ x = \frac{-15 - 9}{2} = \frac{-24}{2} = -12 \]

Now, we have the two zeros of the quadratic function:

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

So, the final result is:

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

Answer 2

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=

Answer 3

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

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

For this function, the coefficients are:

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

Now, let's plug these values into the quadratic formula.

1. Calculate the discriminant: \[ b^2 - 4ac = 16^2 - 4 \cdot (-1) \cdot (-39) = 256 - 156 = 100 \]

2. Use the quadratic formula: \[ x = \frac{-16 \pm \sqrt{100}}{2 \cdot (-1)} = \frac{-16 \pm 10}{-2} \]

3. Calculate the two possible values for \( x \):

   • For the positive case: \[ x = \frac{-16 + 10}{-2} = \frac{-6}{-2} = 3 \]
   • For the negative case: \[ x = \frac{-16 - 10}{-2} = \frac{-26}{-2} = 13 \]

Therefore, the zeros of the quadratic function are:

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

So the final answers are:

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

Answer 4

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

Answer 5

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

1. Axis of symmetry: The formula for the axis of symmetry for a quadratic function \( y = ax^2 + bx + c \) is given by: \[ x = -\frac{b}{2a} \]

2. Vertex: Once we have the axis of symmetry \( x \), we can find the vertex by substituting this value back into the original function to find the \( y \)-coordinate.

Step 1: Calculate the axis of symmetry

For the given function:

• \( a = 1 \)
• \( b = -10 \)

Using the formula: \[ x = -\frac{-10}{2 \cdot 1} = \frac{10}{2} = 5 \]

Step 2: Calculate the vertex

Now, substitute \( x = 5 \) back into the function to find the \( y \)-coordinate of the vertex: \[ y = (5)^2 - 10(5) + 3 \] \[ y = 25 - 50 + 3 \] \[ y = -22 \]

Conclusion

Thus, the vertex is at \( (5, -22) \) and the axis of symmetry is \( x = 5 \).

Final answer:

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

Answer 6

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

Answer 7

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

1. Identify the vertex and the axis of symmetry of the quadratic function.
2. Find the corresponding point on the graph for the x-coordinate of the given point.
3. Use this point to find the symmetrical point.

Step 1: Find the vertex and axis of symmetry

The quadratic function is in the form \(y = ax^2 + bx + c\), where:

• \(a = 1\)
• \(b = 12\)
• \(c = 8\)

The axis of symmetry is given by: \[ x = -\frac{b}{2a} = -\frac{12}{2 \cdot 1} = -6 \]

To find the vertex, substitute \(x = -6\) into the function: \[ y = (-6)^2 + 12(-6) + 8 \] \[ y = 36 - 72 + 8 \] \[ y = -28 \] Thus, the vertex of the parabola is at the point \((-6, -28)\).

Step 2: Find the point on the graph

Now, we want the point on the parabola that is directly above or below the given point \((-1, -3)\).

1. Find the x-coordinate of the line of symmetry of the point \((-1, -3)\): The symmetrical point across the axis of symmetry will have its x-coordinate equal to: \[ x' = -6 + (-6 - (-1)) = -6 + (-5) = -11 \]

Step 3: Find the y-coordinate of the symmetrical point

Now, substitute \(x' = -11\) back into the quadratic to find the corresponding y-coordinate: \[ y' = (-11)^2 + 12(-11) + 8 \] \[ y' = 121 - 132 + 8 \] \[ y' = -3 \]

Result

The symmetrical point to \((-1, -3)\) with respect to the graph of \(y = x^2 + 12x + 8\) is: \[ \boxed{(-11, -3)} \]

Answer 8

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: (
)

Answer 9

To find the key points on the graph 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 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 this function, we have:

• \( a = 1 \)
• \( b = -16 \)

Calculating the x-coordinate of the vertex:

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

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

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

Thus, the vertex is at:

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

Step 2: Find the x-intercepts

The x-intercepts occur when \( y = 0 \): \[ 0 = x^2 - 16x + 48 \]

We can factor the quadratic equation:

\[ x^2 - 16x + 48 = (x - 12)(x - 4) = 0 \]

Setting each factor equal to zero gives us the x-intercepts:

\[ x - 12 = 0 \quad \Rightarrow \quad x = 12 \] \[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]

Thus, the x-intercepts are:

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

Step 3: Find the y-intercept

The y-intercept occurs when \( x = 0 \):

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

Thus, the y-intercept is:

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

Summary of Key Points

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

So the final results are:

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