Question
Question 1 A)Determine the zeros of the quadratic function y=x2+15x+36y=x2+15x+36. (When listing the smaller zero, remember that negative values are always smaller than positive values.)(1 point) smaller zero: x=x=$$; greater zero: x=x=$$ Question 2 A)Determine the zeros of the quadratic function y=−x2+16x−39y=−x2+16x−39. (When listing the smaller zero, remember that negative values are always smaller than positive values.)(1 point) smaller zero: x=x=$$; greater zero: x=x=$$ Question 3 A)Determine the vertex and axis of symmetry of y=x2−10x+3y=x2−10x+3.(1 point) vertex: ($$); axis of symmetry: x=x=$$ Question 4 A)What point is symmetrical to (−1,−3)(−1,−3) for the graph of y=x2+12x+8y=x2+12x+8?(1 point) ($$) Question 5 A)What are the key points on the graph of y=x2−16x+48y=x2−16x+48? Name the vertex, x-intercepts, and y-intercept.(2 points) vertex: ($$) (smaller) x-intercept: ($$) (larger) x-intercept: ($$) y-intercept: ($$) page 19 of 20
Answers
GPT-4o mini
Let's solve each of the questions step by step.
### Question 1
We need to find the zeros of the quadratic function \( y = x^2 + 15x + 36 \).
To find the zeros, we can factor the quadratic:
\[
x^2 + 15x + 36 = (x + 3)(x + 12) = 0
\]
Setting each factor to zero gives:
\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]
\[
x + 12 = 0 \quad \Rightarrow \quad x = -12
\]
Thus, the zeros are:
- smaller zero: \( x = -12 \)
- greater zero: \( x = -3 \)
### Question 2
Now we find the zeros of the quadratic function \( y = -x^2 + 16x - 39 \).
We can set the equation to zero:
\[
-x^2 + 16x - 39 = 0 \quad \text{(multiply by -1 for simplicity)}
\]
\[
x^2 - 16x + 39 = 0
\]
Next, we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1, b = -16, c = 39 \):
\[
x = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 39}}{2 \cdot 1} = \frac{16 \pm \sqrt{256 - 156}}{2} = \frac{16 \pm \sqrt{100}}{2} = \frac{16 \pm 10}{2}
\]
Calculating the solutions:
\[
x = \frac{26}{2} = 13 \quad \text{(greater zero)}
\]
\[
x = \frac{6}{2} = 3 \quad \text{(smaller zero)}
\]
Thus, the zeros are:
- smaller zero: \( x = 3 \)
- greater zero: \( x = 13 \)
### Question 3
For the vertex and axis of symmetry of \( y = x^2 - 10x + 3 \):
The x-coordinate of the vertex is given by:
\[
x = -\frac{b}{2a} = -\frac{-10}{2 \cdot 1} = \frac{10}{2} = 5
\]
To find the y-coordinate of the vertex, substitute \( x = 5 \) into the equation:
\[
y = (5)^2 - 10(5) + 3 = 25 - 50 + 3 = -22
\]
Thus, the vertex is \( (5, -22) \) and the axis of symmetry is \( x = 5 \).
### Question 4
We need to find the symmetrical point to \( (-1, -3) \) for the graph of \( y=x^2 + 12x + 8 \).
First, we calculate the vertex of the parabola:
\[
x = -\frac{12}{2} = -6
\]
Next, we find the y-coordinate of the vertex:
\[
y = (-6)^2 + 12(-6) + 8 = 36 - 72 + 8 = -28
\]
The vertex is \( (-6, -28) \). The line of symmetry is \( x = -6 \).
To find the point symmetrical to \( (-1, -3) \), we compute the distance from \( -1 \) to \( -6 \):
\[
-6 - (-1) = -5
\]
We move the same distance to the left of \( -6 \):
\[
-6 - 5 = -11
\]
So the symmetrical point is \( (-11, -3) \).
### Question 5
We find the key points for the graph of \( y = x^2 - 16x + 48 \).
1. **Vertex**: The x-coordinate is:
\[
x = \frac{16}{2} = 8
\]
Now substitute \( x = 8 \) into the function:
\[
y = (8)^2 - 16(8) + 48 = 64 - 128 + 48 = -16
\]
So, the vertex is \( (8, -16) \).
2. **X-intercepts**: Set \( y = 0 \):
\[
x^2 - 16x + 48 = 0
\]
Factoring:
\[
(x - 12)(x - 4) = 0
\]
This gives \( x = 12 \) and \( x = 4 \).
3. **Y-intercept**: Set \( x = 0 \):
\[
y = 0^2 - 16(0) + 48 = 48
\]
Thus the y-intercept is \( (0, 48) \).
### Final Answers
- **Question 1**: smaller zero: \( x = -12 \); greater zero: \( x = -3 \)
- **Question 2**: smaller zero: \( x = 3 \); greater zero: \( x = 13 \)
- **Question 3**: vertex: \( (5, -22) \); axis of symmetry: \( x = 5 \)
- **Question 4**: point: \( (-11, -3) \)
- **Question 5**: vertex: \( (8, -16) \); smaller x-intercept: \( (4, 0) \); larger x-intercept: \( (12, 0) \); y-intercept: \( (0, 48) \)
### Question 1
We need to find the zeros of the quadratic function \( y = x^2 + 15x + 36 \).
To find the zeros, we can factor the quadratic:
\[
x^2 + 15x + 36 = (x + 3)(x + 12) = 0
\]
Setting each factor to zero gives:
\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]
\[
x + 12 = 0 \quad \Rightarrow \quad x = -12
\]
Thus, the zeros are:
- smaller zero: \( x = -12 \)
- greater zero: \( x = -3 \)
### Question 2
Now we find the zeros of the quadratic function \( y = -x^2 + 16x - 39 \).
We can set the equation to zero:
\[
-x^2 + 16x - 39 = 0 \quad \text{(multiply by -1 for simplicity)}
\]
\[
x^2 - 16x + 39 = 0
\]
Next, we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1, b = -16, c = 39 \):
\[
x = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 39}}{2 \cdot 1} = \frac{16 \pm \sqrt{256 - 156}}{2} = \frac{16 \pm \sqrt{100}}{2} = \frac{16 \pm 10}{2}
\]
Calculating the solutions:
\[
x = \frac{26}{2} = 13 \quad \text{(greater zero)}
\]
\[
x = \frac{6}{2} = 3 \quad \text{(smaller zero)}
\]
Thus, the zeros are:
- smaller zero: \( x = 3 \)
- greater zero: \( x = 13 \)
### Question 3
For the vertex and axis of symmetry of \( y = x^2 - 10x + 3 \):
The x-coordinate of the vertex is given by:
\[
x = -\frac{b}{2a} = -\frac{-10}{2 \cdot 1} = \frac{10}{2} = 5
\]
To find the y-coordinate of the vertex, substitute \( x = 5 \) into the equation:
\[
y = (5)^2 - 10(5) + 3 = 25 - 50 + 3 = -22
\]
Thus, the vertex is \( (5, -22) \) and the axis of symmetry is \( x = 5 \).
### Question 4
We need to find the symmetrical point to \( (-1, -3) \) for the graph of \( y=x^2 + 12x + 8 \).
First, we calculate the vertex of the parabola:
\[
x = -\frac{12}{2} = -6
\]
Next, we find the y-coordinate of the vertex:
\[
y = (-6)^2 + 12(-6) + 8 = 36 - 72 + 8 = -28
\]
The vertex is \( (-6, -28) \). The line of symmetry is \( x = -6 \).
To find the point symmetrical to \( (-1, -3) \), we compute the distance from \( -1 \) to \( -6 \):
\[
-6 - (-1) = -5
\]
We move the same distance to the left of \( -6 \):
\[
-6 - 5 = -11
\]
So the symmetrical point is \( (-11, -3) \).
### Question 5
We find the key points for the graph of \( y = x^2 - 16x + 48 \).
1. **Vertex**: The x-coordinate is:
\[
x = \frac{16}{2} = 8
\]
Now substitute \( x = 8 \) into the function:
\[
y = (8)^2 - 16(8) + 48 = 64 - 128 + 48 = -16
\]
So, the vertex is \( (8, -16) \).
2. **X-intercepts**: Set \( y = 0 \):
\[
x^2 - 16x + 48 = 0
\]
Factoring:
\[
(x - 12)(x - 4) = 0
\]
This gives \( x = 12 \) and \( x = 4 \).
3. **Y-intercept**: Set \( x = 0 \):
\[
y = 0^2 - 16(0) + 48 = 48
\]
Thus the y-intercept is \( (0, 48) \).
### Final Answers
- **Question 1**: smaller zero: \( x = -12 \); greater zero: \( x = -3 \)
- **Question 2**: smaller zero: \( x = 3 \); greater zero: \( x = 13 \)
- **Question 3**: vertex: \( (5, -22) \); axis of symmetry: \( x = 5 \)
- **Question 4**: point: \( (-11, -3) \)
- **Question 5**: vertex: \( (8, -16) \); smaller x-intercept: \( (4, 0) \); larger x-intercept: \( (12, 0) \); y-intercept: \( (0, 48) \)