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Determine the zeros of the quadratic function y=−x2+16x−39 . (When listing the smaller zero, remember that negative values are...Question
Determine the zeros of the quadratic function y=x^2 + 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= ___
smaller zero: x= ___
; greater zero: x= ___
Answers
Answer
Determine the zeros of the quadratic function y=−x^2 + 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= ___
Answer
Determine the vertex and axis of symmetry of y=x^2 − 10x + 3
.(1 point)
vertex: ( ___ ); axis of symmetry: x= ___
.(1 point)
vertex: ( ___ ); axis of symmetry: x= ___
Answer
What point is symmetrical to (−1,−3)
for the graph of y=x^2 + 12x + 8
?(1 point)
( ___ )
for the graph of y=x^2 + 12x + 8
?(1 point)
( ___ )
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 zeros of the quadratic function \( y = x^2 + 15x + 36 \), we need to solve the equation \( x^2 + 15x + 36 = 0 \).
We can use the quadratic formula, which is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = 15 \), and \( c = 36 \).
First, calculate the discriminant:
\[
b^2 - 4ac = 15^2 - 4(1)(36) = 225 - 144 = 81
\]
Now, we can plug the values into the quadratic formula:
\[
x = \frac{-15 \pm \sqrt{81}}{2(1)} = \frac{-15 \pm 9}{2}
\]
This gives us two solutions:
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 function are:
- Smaller zero: \( x = -12 \)
- Greater zero: \( x = -3 \)
Final answer:
- smaller zero: \( x = -12 \)
- greater zero: \( x = -3 \)
We can use the quadratic formula, which is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = 15 \), and \( c = 36 \).
First, calculate the discriminant:
\[
b^2 - 4ac = 15^2 - 4(1)(36) = 225 - 144 = 81
\]
Now, we can plug the values into the quadratic formula:
\[
x = \frac{-15 \pm \sqrt{81}}{2(1)} = \frac{-15 \pm 9}{2}
\]
This gives us two solutions:
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 function are:
- Smaller zero: \( x = -12 \)
- Greater zero: \( x = -3 \)
Final answer:
- smaller zero: \( x = -12 \)
- greater zero: \( x = -3 \)
Answered by
GPT-4o mini
To determine the zeros of the quadratic function \( y = -x^2 + 16x - 39 \), we need to solve the equation:
\[
-x^2 + 16x - 39 = 0
\]
We can rearrange this to:
\[
x^2 - 16x + 39 = 0
\]
Now, we will apply 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:
\[
b^2 - 4ac = (-16)^2 - 4(1)(39) = 256 - 156 = 100
\]
Now we can use the quadratic formula:
\[
x = \frac{-(-16) \pm \sqrt{100}}{2(1)} = \frac{16 \pm 10}{2}
\]
This gives us two solutions:
1. For the positive case:
\[
x = \frac{16 + 10}{2} = \frac{26}{2} = 13
\]
2. For the negative case:
\[
x = \frac{16 - 10}{2} = \frac{6}{2} = 3
\]
Thus, the zeros of the function are:
- Smaller zero: \( x = 3 \)
- Greater zero: \( x = 13 \)
Final answer:
- smaller zero: \( x = 3 \)
- greater zero: \( x = 13 \)
\[
-x^2 + 16x - 39 = 0
\]
We can rearrange this to:
\[
x^2 - 16x + 39 = 0
\]
Now, we will apply 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:
\[
b^2 - 4ac = (-16)^2 - 4(1)(39) = 256 - 156 = 100
\]
Now we can use the quadratic formula:
\[
x = \frac{-(-16) \pm \sqrt{100}}{2(1)} = \frac{16 \pm 10}{2}
\]
This gives us two solutions:
1. For the positive case:
\[
x = \frac{16 + 10}{2} = \frac{26}{2} = 13
\]
2. For the negative case:
\[
x = \frac{16 - 10}{2} = \frac{6}{2} = 3
\]
Thus, the zeros of the function are:
- Smaller zero: \( x = 3 \)
- Greater zero: \( x = 13 \)
Final answer:
- smaller zero: \( x = 3 \)
- greater zero: \( x = 13 \)
Answered by
GPT-4o mini
To find the vertex and axis of symmetry of the quadratic function \( y = x^2 - 10x + 3 \), we can use the vertex formula and the property of the axis of symmetry.
The vertex \( (h, k) \) of a quadratic in the form \( y = ax^2 + bx + c \) can be found using the formulas:
- The x-coordinate of the vertex \( h \) is given by:
\[
h = -\frac{b}{2a}
\]
- The y-coordinate of the vertex \( k \) can be found by substituting \( h \) back into the quadratic function.
In our case:
- \( a = 1 \)
- \( b = -10 \)
- \( c = 3 \)
1. **Calculate \( h \)**:
\[
h = -\frac{-10}{2 \cdot 1} = \frac{10}{2} = 5
\]
2. **Calculate \( k \)** by substituting \( h \) into the function:
\[
k = (5)^2 - 10(5) + 3 = 25 - 50 + 3 = -22
\]
Thus, the vertex is \( (5, -22) \).
3. **Find the axis of symmetry**:
The axis of symmetry is given by the line \( x = h \).
So in this case:
\[
\text{axis of symmetry} : x = 5
\]
Final answer:
- vertex: \( (5, -22) \)
- axis of symmetry: \( x = 5 \)
The vertex \( (h, k) \) of a quadratic in the form \( y = ax^2 + bx + c \) can be found using the formulas:
- The x-coordinate of the vertex \( h \) is given by:
\[
h = -\frac{b}{2a}
\]
- The y-coordinate of the vertex \( k \) can be found by substituting \( h \) back into the quadratic function.
In our case:
- \( a = 1 \)
- \( b = -10 \)
- \( c = 3 \)
1. **Calculate \( h \)**:
\[
h = -\frac{-10}{2 \cdot 1} = \frac{10}{2} = 5
\]
2. **Calculate \( k \)** by substituting \( h \) into the function:
\[
k = (5)^2 - 10(5) + 3 = 25 - 50 + 3 = -22
\]
Thus, the vertex is \( (5, -22) \).
3. **Find the axis of symmetry**:
The axis of symmetry is given by the line \( x = h \).
So in this case:
\[
\text{axis of symmetry} : x = 5
\]
Final answer:
- vertex: \( (5, -22) \)
- axis of symmetry: \( x = 5 \)
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 need to follow these steps:
1. **Find the vertex of the quadratic**: The vertex of a parabola in the form \(y = ax^2 + bx + c\) can be found using the formula for the x-coordinate of the vertex:
\[
h = -\frac{b}{2a}
\]
In this case:
- \(a = 1\)
- \(b = 12\)
Therefore,
\[
h = -\frac{12}{2 \cdot 1} = -\frac{12}{2} = -6
\]
To find the y-coordinate of the vertex \(k\), substitute \(h\) back into the equation:
\[
k = (-6)^2 + 12(-6) + 8 = 36 - 72 + 8 = -28
\]
Thus, the vertex of the parabola is \((-6, -28)\).
2. **Find the line of symmetry**: The line of symmetry for the parabola is \(x = -6\).
3. **Determine the symmetrical point**: The point \((-1, -3)\) is on the left side of the line of symmetry \(x = -6\). To find its symmetrical point, we need to find the distance from \(-1\) to \(-6\):
\[
\text{Distance} = -6 - (-1) = -6 + 1 = -5
\]
The symmetrical point will be \(5\) units to the right of \(-6\):
\[
x_{\text{symmetrical}} = -6 + 5 = -1
\]
The y-coordinate remains the same since symmetry across the vertex leads to mirroring vertically at the vertex level, so we calculate the y-value of the original point and then reflect over the vertex.
4. **Find the y-coordinate of the symmetrical point**:
Substitute \(x = -1\) into the original equation to find \(y\):
\[
y = (-1)^2 + 12(-1) + 8 = 1 - 12 + 8 = -3
\]
Thus, the symmetrical point of \((-1, -3)\) across the vertex (\(-6, -28\)) is at the same vertical height as the original:
Therefore, the symmetrical point to \((-1, -3)\) is:
\[
(-11, -3)
\]
So, the answer is:
( \( -11, -3 \) )
1. **Find the vertex of the quadratic**: The vertex of a parabola in the form \(y = ax^2 + bx + c\) can be found using the formula for the x-coordinate of the vertex:
\[
h = -\frac{b}{2a}
\]
In this case:
- \(a = 1\)
- \(b = 12\)
Therefore,
\[
h = -\frac{12}{2 \cdot 1} = -\frac{12}{2} = -6
\]
To find the y-coordinate of the vertex \(k\), substitute \(h\) back into the equation:
\[
k = (-6)^2 + 12(-6) + 8 = 36 - 72 + 8 = -28
\]
Thus, the vertex of the parabola is \((-6, -28)\).
2. **Find the line of symmetry**: The line of symmetry for the parabola is \(x = -6\).
3. **Determine the symmetrical point**: The point \((-1, -3)\) is on the left side of the line of symmetry \(x = -6\). To find its symmetrical point, we need to find the distance from \(-1\) to \(-6\):
\[
\text{Distance} = -6 - (-1) = -6 + 1 = -5
\]
The symmetrical point will be \(5\) units to the right of \(-6\):
\[
x_{\text{symmetrical}} = -6 + 5 = -1
\]
The y-coordinate remains the same since symmetry across the vertex leads to mirroring vertically at the vertex level, so we calculate the y-value of the original point and then reflect over the vertex.
4. **Find the y-coordinate of the symmetrical point**:
Substitute \(x = -1\) into the original equation to find \(y\):
\[
y = (-1)^2 + 12(-1) + 8 = 1 - 12 + 8 = -3
\]
Thus, the symmetrical point of \((-1, -3)\) across the vertex (\(-6, -28\)) is at the same vertical height as the original:
Therefore, the symmetrical point to \((-1, -3)\) is:
\[
(-11, -3)
\]
So, the answer is:
( \( -11, -3 \) )
Answered by
GPT-4o mini
To find the key points on the graph of the quadratic function \( y = x^2 - 16x + 48 \), we will identify the vertex, the x-intercepts, and the y-intercept.
### Vertex
The vertex of a quadratic function in the form \( y = ax^2 + bx + c\) can be found with the following formulas:
1. **Calculate the x-coordinate of the vertex \( h \)**:
\[
h = -\frac{b}{2a}
\]
In this case:
- \( a = 1 \)
- \( b = -16 \)
Therefore:
\[
h = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8
\]
2. **Calculate the y-coordinate \( k \)** by substituting \( h \) back into the function:
\[
k = 8^2 - 16(8) + 48 = 64 - 128 + 48 = -16
\]
Thus, the vertex is:
\[
\text{vertex: } (8, -16)
\]
### X-Intercepts
The x-intercepts occur when \( y = 0 \):
\[
x^2 - 16x + 48 = 0
\]
We can solve this using factoring:
\[
(x - 12)(x - 4) = 0
\]
Setting each factor to zero gives us:
\[
x - 12 = 0 \quad \Rightarrow \quad x = 12
\]
\[
x - 4 = 0 \quad \Rightarrow \quad x = 4
\]
So the x-intercepts are:
- Smaller x-intercept: \( (4, 0) \)
- Larger x-intercept: \( (12, 0) \)
### Y-Intercept
The y-intercept occurs when \( x = 0 \):
\[
y = 0^2 - 16(0) + 48 = 48
\]
So 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**: \( (0, 48) \)
Final Answer:
- vertex: \( (8, -16) \)
- (smaller) x-intercept: \( (4, 0) \)
- (larger) x-intercept: \( (12, 0) \)
- y-intercept: \( (0, 48) \)
### Vertex
The vertex of a quadratic function in the form \( y = ax^2 + bx + c\) can be found with the following formulas:
1. **Calculate the x-coordinate of the vertex \( h \)**:
\[
h = -\frac{b}{2a}
\]
In this case:
- \( a = 1 \)
- \( b = -16 \)
Therefore:
\[
h = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8
\]
2. **Calculate the y-coordinate \( k \)** by substituting \( h \) back into the function:
\[
k = 8^2 - 16(8) + 48 = 64 - 128 + 48 = -16
\]
Thus, the vertex is:
\[
\text{vertex: } (8, -16)
\]
### X-Intercepts
The x-intercepts occur when \( y = 0 \):
\[
x^2 - 16x + 48 = 0
\]
We can solve this using factoring:
\[
(x - 12)(x - 4) = 0
\]
Setting each factor to zero gives us:
\[
x - 12 = 0 \quad \Rightarrow \quad x = 12
\]
\[
x - 4 = 0 \quad \Rightarrow \quad x = 4
\]
So the x-intercepts are:
- Smaller x-intercept: \( (4, 0) \)
- Larger x-intercept: \( (12, 0) \)
### Y-Intercept
The y-intercept occurs when \( x = 0 \):
\[
y = 0^2 - 16(0) + 48 = 48
\]
So 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**: \( (0, 48) \)
Final Answer:
- vertex: \( (8, -16) \)
- (smaller) x-intercept: \( (4, 0) \)
- (larger) x-intercept: \( (12, 0) \)
- y-intercept: \( (0, 48) \)
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