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To solve the problem of identifying which function has an [tex]$x$[/tex]-intercept of 4, we need to consider the definition of an [tex]$x$[/tex]-intercept. An [tex]$x$[/tex]-intercept occurs where the function crosses the x-axis, meaning the value of the function (f(x)) is 0 at that point. Therefore, we need to find which function satisfies [tex]$f(4) = 0$[/tex].
Let’s look at each function provided in the options:
Option A: [tex]\( f(x) = -\frac{1}{2}x + 4 \)[/tex]
To test if this function has an [tex]$x$[/tex]-intercept at 4, substitute [tex]\( x = 4 \)[/tex] into the function:
[tex]\[ f(4) = -\frac{1}{2}(4) + 4 \][/tex]
[tex]\[ f(4) = -2 + 4 \][/tex]
[tex]\[ f(4) = 2 \neq 0 \][/tex]
Since [tex]\( f(4) = 2 \)[/tex], this function does not have an [tex]$x$[/tex]-intercept at [tex]\( x = 4 \)[/tex].
Option B: [tex]\( f(x) = (x + 4)(x - 7) \)[/tex]
To test if this function has an [tex]$x$[/tex]-intercept at 4, substitute [tex]\( x = 4 \)[/tex] into the function:
[tex]\[ f(4) = (4 + 4)(4 - 7) \][/tex]
[tex]\[ f(4) = 8 \times -3 \][/tex]
[tex]\[ f(4) = -24 \neq 0 \][/tex]
Since [tex]\( f(4) = -24 \)[/tex], this function does not have an [tex]$x$[/tex]-intercept at [tex]\( x = 4 \)[/tex].
Option C: [tex]\( f(x) = x^2 + 3x - 4 \)[/tex]
To find the [tex]$x$[/tex]-intercepts of this quadratic function, we need to solve for when [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ 0 = x^2 + 3x - 4 \][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = -4 \)[/tex]:
[tex]\[ x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-4)}}{2(1)} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{9 + 16}}{2} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{25}}{2} \][/tex]
[tex]\[ x = \frac{-3 \pm 5}{2} \][/tex]
[tex]\[ x = 1 \quad \text{or} \quad x = -4 \][/tex]
Since the [tex]$x$[/tex]-intercepts are [tex]\( x = 1 \)[/tex] and [tex]\( x = -4 \)[/tex], the function does not have an [tex]$x$[/tex]-intercept at [tex]\( x = 4 \)[/tex].
Option D: [tex]\( f(x) = x - 4 \)[/tex]
To test if this function has an [tex]$x$[/tex]-intercept at 4, substitute [tex]\( x = 4 \)[/tex] into the function:
[tex]\[ f(4) = 4 - 4 \][/tex]
[tex]\[ f(4) = 0 \][/tex]
Since [tex]\( f(4) = 0 \)[/tex], this function does have an [tex]$x$[/tex]-intercept at [tex]\( x = 4 \)[/tex].
Therefore, the correct answer is:
d. D
Let’s look at each function provided in the options:
Option A: [tex]\( f(x) = -\frac{1}{2}x + 4 \)[/tex]
To test if this function has an [tex]$x$[/tex]-intercept at 4, substitute [tex]\( x = 4 \)[/tex] into the function:
[tex]\[ f(4) = -\frac{1}{2}(4) + 4 \][/tex]
[tex]\[ f(4) = -2 + 4 \][/tex]
[tex]\[ f(4) = 2 \neq 0 \][/tex]
Since [tex]\( f(4) = 2 \)[/tex], this function does not have an [tex]$x$[/tex]-intercept at [tex]\( x = 4 \)[/tex].
Option B: [tex]\( f(x) = (x + 4)(x - 7) \)[/tex]
To test if this function has an [tex]$x$[/tex]-intercept at 4, substitute [tex]\( x = 4 \)[/tex] into the function:
[tex]\[ f(4) = (4 + 4)(4 - 7) \][/tex]
[tex]\[ f(4) = 8 \times -3 \][/tex]
[tex]\[ f(4) = -24 \neq 0 \][/tex]
Since [tex]\( f(4) = -24 \)[/tex], this function does not have an [tex]$x$[/tex]-intercept at [tex]\( x = 4 \)[/tex].
Option C: [tex]\( f(x) = x^2 + 3x - 4 \)[/tex]
To find the [tex]$x$[/tex]-intercepts of this quadratic function, we need to solve for when [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ 0 = x^2 + 3x - 4 \][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = -4 \)[/tex]:
[tex]\[ x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-4)}}{2(1)} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{9 + 16}}{2} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{25}}{2} \][/tex]
[tex]\[ x = \frac{-3 \pm 5}{2} \][/tex]
[tex]\[ x = 1 \quad \text{or} \quad x = -4 \][/tex]
Since the [tex]$x$[/tex]-intercepts are [tex]\( x = 1 \)[/tex] and [tex]\( x = -4 \)[/tex], the function does not have an [tex]$x$[/tex]-intercept at [tex]\( x = 4 \)[/tex].
Option D: [tex]\( f(x) = x - 4 \)[/tex]
To test if this function has an [tex]$x$[/tex]-intercept at 4, substitute [tex]\( x = 4 \)[/tex] into the function:
[tex]\[ f(4) = 4 - 4 \][/tex]
[tex]\[ f(4) = 0 \][/tex]
Since [tex]\( f(4) = 0 \)[/tex], this function does have an [tex]$x$[/tex]-intercept at [tex]\( x = 4 \)[/tex].
Therefore, the correct answer is:
d. D
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