Question
etermine 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)
smaller zero: x=
; greater zero: x=
. (When listing the smaller zero, remember that negative values are always smaller than positive values.)(1 point)
smaller zero: x=
; greater zero: x=
Answers
Answered by
GPT-4o mini
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: \( x = -3 \)
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
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
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=
. (When listing the smaller zero, remember that negative values are always smaller than positive values.)(1 point)
smaller zero: x=
; greater zero: x=
Answered by
GPT-4o mini
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: \( x = 13 \)
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
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
Determine the vertex and axis of symmetry of y=x2−10x+3
.(1 point)
vertex: (
); axis of symmetry: x=
.(1 point)
vertex: (
); axis of symmetry: x=
Answered by
GPT-4o mini
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: \( x = 5 \)
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
What point is symmetrical to (−1,−3)
for the graph of y=x2+12x+8
?(1 point)
(
)
for the graph of y=x2+12x+8
?(1 point)
(
)
Answered by
GPT-4o mini
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:
\[
\boxed{(-11, -3)}
\]
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
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: (
)
? Name the vertex, x-intercepts, and y-intercept.(2 points)
vertex: (
)
(smaller) x-intercept: (
)
(larger) x-intercept: (
)
y-intercept: (
)
Answered by
GPT-4o mini
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: \( (0, 48) \)
### 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: \( (0, 48) \)