Question
5.
Determine the value of the constant term of the quadratic function in standard form, given its graph.
(1 point)
The value of c in the standard form of the quadratic function y = ax2+bx+c
is
Question 8
6.
In the xy-coordinate plane, the graph of the equations y = 3x2 −12x −36
has zeros at x = and x =,
where < .
The graph has a minimum at ( , -48).
(3 points)
What are the values of
=
=
and
=
Determine the value of the constant term of the quadratic function in standard form, given its graph.
(1 point)
The value of c in the standard form of the quadratic function y = ax2+bx+c
is
Question 8
6.
In the xy-coordinate plane, the graph of the equations y = 3x2 −12x −36
has zeros at x = and x =,
where < .
The graph has a minimum at ( , -48).
(3 points)
What are the values of
=
=
and
=
Answers
Answer
5.
Determine the value of the constant term of the quadratic function in standard form, given its graph.
(1 point)
The value of c in the standard form of the quadratic function y = ax2+bx+c
is
Determine the value of the constant term of the quadratic function in standard form, given its graph.
(1 point)
The value of c in the standard form of the quadratic function y = ax2+bx+c
is
Answered by
GPT-4o mini
To answer your questions, let's break them down step by step.
### Question 5
The constant term \( c \) in the standard form of the quadratic function \( y = ax^2 + bx + c \) is the value of \( y \) when \( x = 0 \). This term affects the vertical position of the graph.
Since we don't have the graph to provide specific values, if you have it, simply identify the \( y \)-intercept on the graph to find \( c \).
### Question 6
To find the zeros of the quadratic function \( y = 3x^2 - 12x - 36 \), we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where:
- \( a = 3 \)
- \( b = -12 \)
- \( c = -36 \)
1. Calculate the discriminant:
\[
b^2 - 4ac = (-12)^2 - 4(3)(-36) = 144 + 432 = 576
\]
2. Now find the zeros:
\[
x = \frac{-(-12) \pm \sqrt{576}}{2(3)} = \frac{12 \pm 24}{6}
\]
This gives us two values:
- \( x_1 = \frac{12 + 24}{6} = \frac{36}{6} = 6 \)
- \( x_2 = \frac{12 - 24}{6} = \frac{-12}{6} = -2 \)
So, the zeros are:
- \( x = -2 \)
- \( x = 6 \)
### Minimum Point
To find the vertex (minimum) of the quadratic function, we can use the vertex formula:
\[
x = -\frac{b}{2a}
\]
For our function:
\[
x = -\frac{-12}{2(3)} = \frac{12}{6} = 2
\]
Now substitute \( x = 2 \) back into the function to find the \( y \)-coordinate:
\[
y = 3(2)^2 - 12(2) - 36 = 3(4) - 24 - 36 = 12 - 24 - 36 = -48
\]
Thus, the vertex is at \( (2, -48) \).
### Final Answers
1. The zeros are:
- \( x = -2 \) and \( x = 6 \).
2. The minimum point is:
- \( (2, -48) \).
So, in the format requested:
- \( x_1 = -2 \)
- \( x_2 = 6 \)
- Minimum point \( (2, -48) \) implies \( = 2 \) and the minimum value (the second coordinate) is \( = -48 \).
### Question 5
The constant term \( c \) in the standard form of the quadratic function \( y = ax^2 + bx + c \) is the value of \( y \) when \( x = 0 \). This term affects the vertical position of the graph.
Since we don't have the graph to provide specific values, if you have it, simply identify the \( y \)-intercept on the graph to find \( c \).
### Question 6
To find the zeros of the quadratic function \( y = 3x^2 - 12x - 36 \), we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where:
- \( a = 3 \)
- \( b = -12 \)
- \( c = -36 \)
1. Calculate the discriminant:
\[
b^2 - 4ac = (-12)^2 - 4(3)(-36) = 144 + 432 = 576
\]
2. Now find the zeros:
\[
x = \frac{-(-12) \pm \sqrt{576}}{2(3)} = \frac{12 \pm 24}{6}
\]
This gives us two values:
- \( x_1 = \frac{12 + 24}{6} = \frac{36}{6} = 6 \)
- \( x_2 = \frac{12 - 24}{6} = \frac{-12}{6} = -2 \)
So, the zeros are:
- \( x = -2 \)
- \( x = 6 \)
### Minimum Point
To find the vertex (minimum) of the quadratic function, we can use the vertex formula:
\[
x = -\frac{b}{2a}
\]
For our function:
\[
x = -\frac{-12}{2(3)} = \frac{12}{6} = 2
\]
Now substitute \( x = 2 \) back into the function to find the \( y \)-coordinate:
\[
y = 3(2)^2 - 12(2) - 36 = 3(4) - 24 - 36 = 12 - 24 - 36 = -48
\]
Thus, the vertex is at \( (2, -48) \).
### Final Answers
1. The zeros are:
- \( x = -2 \) and \( x = 6 \).
2. The minimum point is:
- \( (2, -48) \).
So, in the format requested:
- \( x_1 = -2 \)
- \( x_2 = 6 \)
- Minimum point \( (2, -48) \) implies \( = 2 \) and the minimum value (the second coordinate) is \( = -48 \).
Answered by
GPT-4o mini
To determine the value of the constant term \( c \) in the standard form of the quadratic function \( y = ax^2 + bx + c \), you need to look at the graph of the quadratic function. The constant term \( c \) corresponds to the \( y \)-intercept of the graph, which is the point where the graph intersects the \( y \)-axis (where \( x = 0 \)).
### Steps to find \( c \):
1. **Identify the \( y \)-intercept**: Look at the graph of the quadratic equation and find the point where it crosses the \( y \)-axis. This point will have coordinates of the form \( (0, c) \).
2. **Read the \( y \)-value at this point**: The value of \( c \) is simply the \( y \)-coordinate of the intersection point.
If you can provide the specific graph or describe its key points (like the \( y \)-intercept), I can help you find the exact value of \( c \). If you have a graph in mind, please share any relevant information!
### Steps to find \( c \):
1. **Identify the \( y \)-intercept**: Look at the graph of the quadratic equation and find the point where it crosses the \( y \)-axis. This point will have coordinates of the form \( (0, c) \).
2. **Read the \( y \)-value at this point**: The value of \( c \) is simply the \( y \)-coordinate of the intersection point.
If you can provide the specific graph or describe its key points (like the \( y \)-intercept), I can help you find the exact value of \( c \). If you have a graph in mind, please share any relevant information!
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