Question
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Part 1
Consider the function f left parenthesis x right parenthesis equals 2 x squared minus 16 x minus 9.
a. nbsp
Determine, without graphing, whether the function has a minimum value or a maximum value.
b.
Find the minimum or maximum value and determine where it occurs.
c.
Identify the function's domain and its range.
Question content area bottom
Part 1
a. The function has a
▼
value
Part 1
Consider the function f left parenthesis x right parenthesis equals 2 x squared minus 16 x minus 9.
a. nbsp
Determine, without graphing, whether the function has a minimum value or a maximum value.
b.
Find the minimum or maximum value and determine where it occurs.
c.
Identify the function's domain and its range.
Question content area bottom
Part 1
a. The function has a
▼
value
Answers
Answered by
GPT-4o mini
To analyze the function \( f(x) = 2x^2 - 16x - 9 \), we will go through each part of the problem step by step.
### Part a:
To determine whether the function has a minimum value or a maximum value, we can look at the coefficient of the \( x^2 \) term.
- The function is a quadratic in the standard form \( ax^2 + bx + c \), where \( a = 2 \), \( b = -16 \), and \( c = -9 \).
- Since \( a = 2 \) is positive, the parabola opens upwards.
**Conclusion:** The function has a **minimum value**.
### Part b:
To find the minimum or maximum value and where it occurs, we can use the vertex formula of a quadratic function. The x-coordinate of the vertex \( x_v \) is given by:
\[
x_v = -\frac{b}{2a}
\]
Plugging in the values:
\[
x_v = -\frac{-16}{2 \cdot 2} = \frac{16}{4} = 4
\]
Now we will substitute \( x = 4 \) back into the function to find the minimum value \( f(4) \):
\[
f(4) = 2(4^2) - 16(4) - 9
\]
\[
= 2(16) - 64 - 9
\]
\[
= 32 - 64 - 9
\]
\[
= 32 - 73
\]
\[
= -41
\]
**Conclusion:** The minimum value is \( -41 \), and it occurs at \( x = 4 \).
### Part c:
The domain of a quadratic function is all real numbers, since there are no restrictions on the value of \( x \).
**Domain:** \( (-\infty, \infty) \)
To find the range, since the parabola opens upwards and achieves a minimum value of \( -41 \) at \( x = 4 \), the range starts from this minimum value and goes to positive infinity.
**Range:** \( [-41, \infty) \)
### Summary of Results:
a. The function has a **minimum value**.
b. The minimum value is \( -41 \) at \( x = 4 \).
c. The domain is \( (-\infty, \infty) \) and the range is \( [-41, \infty) \).
### Part a:
To determine whether the function has a minimum value or a maximum value, we can look at the coefficient of the \( x^2 \) term.
- The function is a quadratic in the standard form \( ax^2 + bx + c \), where \( a = 2 \), \( b = -16 \), and \( c = -9 \).
- Since \( a = 2 \) is positive, the parabola opens upwards.
**Conclusion:** The function has a **minimum value**.
### Part b:
To find the minimum or maximum value and where it occurs, we can use the vertex formula of a quadratic function. The x-coordinate of the vertex \( x_v \) is given by:
\[
x_v = -\frac{b}{2a}
\]
Plugging in the values:
\[
x_v = -\frac{-16}{2 \cdot 2} = \frac{16}{4} = 4
\]
Now we will substitute \( x = 4 \) back into the function to find the minimum value \( f(4) \):
\[
f(4) = 2(4^2) - 16(4) - 9
\]
\[
= 2(16) - 64 - 9
\]
\[
= 32 - 64 - 9
\]
\[
= 32 - 73
\]
\[
= -41
\]
**Conclusion:** The minimum value is \( -41 \), and it occurs at \( x = 4 \).
### Part c:
The domain of a quadratic function is all real numbers, since there are no restrictions on the value of \( x \).
**Domain:** \( (-\infty, \infty) \)
To find the range, since the parabola opens upwards and achieves a minimum value of \( -41 \) at \( x = 4 \), the range starts from this minimum value and goes to positive infinity.
**Range:** \( [-41, \infty) \)
### Summary of Results:
a. The function has a **minimum value**.
b. The minimum value is \( -41 \) at \( x = 4 \).
c. The domain is \( (-\infty, \infty) \) and the range is \( [-41, \infty) \).