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To determine which function has a maximum and a [tex]\(y\)[/tex]-intercept of 4, let's analyze each function with the given criteria.
### Step 1: Identify the [tex]\(y\)[/tex]-intercept of each function
To find the [tex]\(y\)[/tex]-intercept of a function, we substitute [tex]\(x = 0\)[/tex] into the function.
1. [tex]\(f(x) = 4x^2 + 6x - 1\)[/tex]
[tex]\[ f(0) = 4(0)^2 + 6(0) - 1 = -1 \][/tex]
The [tex]\(y\)[/tex]-intercept is [tex]\(-1\)[/tex].
2. [tex]\(f(x) = -4x^2 + 8x + 5\)[/tex]
[tex]\[ f(0) = -4(0)^2 + 8(0) + 5 = 5 \][/tex]
The [tex]\(y\)[/tex]-intercept is [tex]\(5\)[/tex].
3. [tex]\(f(x) = -x^2 + 2x + 4\)[/tex]
[tex]\[ f(0) = -(0)^2 + 2(0) + 4 = 4 \][/tex]
The [tex]\(y\)[/tex]-intercept is [tex]\(4\)[/tex].
4. [tex]\(f(x) = x^2 + 4x - 4\)[/tex]
[tex]\[ f(0) = (0)^2 + 4(0) - 4 = -4 \][/tex]
The [tex]\(y\)[/tex]-intercept is [tex]\(-4\)[/tex].
From the calculations above, the function [tex]\(f(x) = -x^2 + 2x + 4\)[/tex] is the only function with a [tex]\(y\)[/tex]-intercept of 4.
### Step 2: Determine if the function has a maximum
To determine whether the function has a maximum or minimum, we analyze the leading coefficient of the quadratic term.
- A quadratic function [tex]\(ax^2 + bx + c\)[/tex] has a maximum if [tex]\(a < 0\)[/tex] (negative leading coefficient).
- A quadratic function [tex]\(ax^2 + bx + c\)[/tex] has a minimum if [tex]\(a > 0\)[/tex] (positive leading coefficient).
1. [tex]\(f(x) = 4x^2 + 6x - 1\)[/tex]
- The leading coefficient is [tex]\(4\)[/tex] (positive), so it has a minimum.
2. [tex]\(f(x) = -4x^2 + 8x + 5\)[/tex]
- The leading coefficient is [tex]\(-4\)[/tex] (negative), so it has a maximum.
3. [tex]\(f(x) = -x^2 + 2x + 4\)[/tex]
- The leading coefficient is [tex]\(-1\)[/tex] (negative), so it has a maximum.
4. [tex]\(f(x) = x^2 + 4x - 4\)[/tex]
- The leading coefficient is [tex]\(1\)[/tex] (positive), so it has a minimum.
### Conclusion
The function [tex]\(f(x) = -x^2 + 2x + 4\)[/tex] has a maximum and a [tex]\(y\)[/tex]-intercept of 4.
Thus, the correct function is:
[tex]\[ f(x) = -x^2 + 2x + 4 \][/tex]
### Step 1: Identify the [tex]\(y\)[/tex]-intercept of each function
To find the [tex]\(y\)[/tex]-intercept of a function, we substitute [tex]\(x = 0\)[/tex] into the function.
1. [tex]\(f(x) = 4x^2 + 6x - 1\)[/tex]
[tex]\[ f(0) = 4(0)^2 + 6(0) - 1 = -1 \][/tex]
The [tex]\(y\)[/tex]-intercept is [tex]\(-1\)[/tex].
2. [tex]\(f(x) = -4x^2 + 8x + 5\)[/tex]
[tex]\[ f(0) = -4(0)^2 + 8(0) + 5 = 5 \][/tex]
The [tex]\(y\)[/tex]-intercept is [tex]\(5\)[/tex].
3. [tex]\(f(x) = -x^2 + 2x + 4\)[/tex]
[tex]\[ f(0) = -(0)^2 + 2(0) + 4 = 4 \][/tex]
The [tex]\(y\)[/tex]-intercept is [tex]\(4\)[/tex].
4. [tex]\(f(x) = x^2 + 4x - 4\)[/tex]
[tex]\[ f(0) = (0)^2 + 4(0) - 4 = -4 \][/tex]
The [tex]\(y\)[/tex]-intercept is [tex]\(-4\)[/tex].
From the calculations above, the function [tex]\(f(x) = -x^2 + 2x + 4\)[/tex] is the only function with a [tex]\(y\)[/tex]-intercept of 4.
### Step 2: Determine if the function has a maximum
To determine whether the function has a maximum or minimum, we analyze the leading coefficient of the quadratic term.
- A quadratic function [tex]\(ax^2 + bx + c\)[/tex] has a maximum if [tex]\(a < 0\)[/tex] (negative leading coefficient).
- A quadratic function [tex]\(ax^2 + bx + c\)[/tex] has a minimum if [tex]\(a > 0\)[/tex] (positive leading coefficient).
1. [tex]\(f(x) = 4x^2 + 6x - 1\)[/tex]
- The leading coefficient is [tex]\(4\)[/tex] (positive), so it has a minimum.
2. [tex]\(f(x) = -4x^2 + 8x + 5\)[/tex]
- The leading coefficient is [tex]\(-4\)[/tex] (negative), so it has a maximum.
3. [tex]\(f(x) = -x^2 + 2x + 4\)[/tex]
- The leading coefficient is [tex]\(-1\)[/tex] (negative), so it has a maximum.
4. [tex]\(f(x) = x^2 + 4x - 4\)[/tex]
- The leading coefficient is [tex]\(1\)[/tex] (positive), so it has a minimum.
### Conclusion
The function [tex]\(f(x) = -x^2 + 2x + 4\)[/tex] has a maximum and a [tex]\(y\)[/tex]-intercept of 4.
Thus, the correct function is:
[tex]\[ f(x) = -x^2 + 2x + 4 \][/tex]
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