Question

indicate which of the following four graphs is the correct graph of this function.
$f(x) = 4 - \log_{4}(x + 5)$

Explanation:

Step1: Find the vertical asymptote

For the function \( f(x) = 4-\log_4(x + 5) \), the argument of the logarithm must be positive: \( x+5>0\Rightarrow x>- 5 \). So the vertical asymptote is \( x=-5 \). We check the graphs: the first graph has asymptote \( x = 0 \), the second has \( x=-5 \) (since it approaches a vertical line at \( x=-5 \)), the third has \( x=-5 \)? Wait, no, let's re - check. Wait, the domain is \( x>-5 \), so the vertical asymptote is \( x = - 5 \). Now, let's check the behavior as \( x ightarrow - 5^+ \). As \( x ightarrow - 5^+ \), \( x + 5 ightarrow0^+ \), so \( \log_4(x + 5) ightarrow-\infty \), then \( f(x)=4-\log_4(x + 5) ightarrow4-(-\infty)=\infty \). So as \( x \) approaches \( - 5 \) from the right, the function goes to \( \infty \).

Step2: Check the y - intercept

To find the y - intercept, set \( x = 0 \). Then \( f(0)=4-\log_4(0 + 5)=4-\log_4(5) \). Since \( \log_4(4) = 1 \) and \( \log_4(5)>\log_4(4) = 1 \), so \( f(0)=4-\log_4(5)<4 \). Wait, no, wait \( \log_4(5)=\frac{\ln(5)}{\ln(4)}\approx\frac{1.6094}{1.3863}\approx1.161 \), so \( f(0)=4 - 1.161 = 2.839 \), which is less than 4. Wait, maybe I made a mistake. Wait, let's re - analyze the function. The parent function is \( y=\log_4(x) \), then \( y =-\log_4(x) \) is a reflection over the x - axis, \( y=-\log_4(x + 5) \) is a horizontal shift left by 5 units, and \( y = 4-\log_4(x + 5) \) is a vertical shift up by 4 units.

Now, let's check the four graphs:

- First graph: vertical asymptote at \( x = 0 \), which is wrong (our asymptote is \( x=-5 \)).
- Second graph: vertical asymptote at \( x=-5 \), and as \( x ightarrow - 5^+ \), the function goes up (since \( f(x) ightarrow\infty \)), and when \( x = 0 \), let's see the y - value. Wait, maybe another approach. Let's find a point. Let \( x+5 = 4\Rightarrow x=-1 \), then \( f(-1)=4-\log_4(4)=4 - 1=3 \). So when \( x=-1 \), \( y = 3 \).

Looking at the graphs:

- The second graph (top - right): when \( x=-1 \), let's see the grid. The x - axis has grid lines at - 8, - 4, 0, 4, 8. The y - axis at 0, 4, 8. Wait, maybe the third graph? Wait, no, let's re - check the asymptote and the direction.

Wait, the function \( y =-\log_4(x + 5)+4 \). The parent function \( y=\log_4(x) \) is increasing. \( y =-\log_4(x) \) is decreasing. \( y=-\log_4(x + 5) \) is decreasing (since it's a horizontal shift of a decreasing function) and \( y = 4-\log_4(x + 5) \) is also decreasing. So the function is decreasing.

Now, check the vertical asymptote: \( x=-5 \). So the graph should have a vertical asymptote at \( x=-5 \) and be decreasing.

First graph: asymptote at \( x = 0 \), decreasing? No, it's on the right of \( x = 0 \), but asymptote is wrong.

Second graph: asymptote at \( x=-5 \), and the graph is increasing (since as \( x \) increases, \( y \) increases), which is wrong because our function is decreasing (since the coefficient of \( \log \) is negative).

Third graph (bottom - left): asymptote at \( x=-5 \) (since it's near \( x=-5 \)), and the function is decreasing (as \( x \) increases, \( y \) decreases), and when \( x=-1 \), \( y = 3 \) (which is on the graph).

Fourth graph (bottom - right): asymptote at \( x=-5 \), but as \( x ightarrow - 5^+ \), the function goes down ( \( y ightarrow-\infty \) ), which is wrong because we know \( f(x) ightarrow\infty \) as \( x ightarrow - 5^+ \).

Wait, maybe I messed up the reflection. Wait, \( f(x)=4-\log_4(x + 5)=4+(-\log_4(x + 5)) \). The function \( y =-\log_4(x + 5) \) is a reflection of \( y=\log_4(x + 5) \) over the x - axis. \( y=\log_4(x + 5) \) is increasing, so \( y =-\log_4(x + 5) \) is decreasing, and \( y = 4-\log_4(x + 5) \) is decreasing. So the function is decreasing.

Now, vertical asymptote at \( x=-5 \), decreasing, and when \( x=-1 \), \( y = 3 \), when \( x = 0 \), \( y=4-\log_4(5)\approx2.84 \).

Looking at the bottom - left graph (third graph): it has vertical asymptote at \( x=-5 \), is decreasing, and passes through points that make sense. Wait, maybe the second graph is increasing, which is wrong. The first graph has asymptote at \( x = 0 \), wrong. The fourth graph goes down as \( x ightarrow - 5^+ \), wrong. So the correct graph should be the bottom - left graph? Wait, no, the user provided four graphs: top - left, top - right, bottom - left, bottom - right.

Wait, let's re - calculate the y - intercept: \( x = 0 \), \( f(0)=4-\log_4(5)\approx4 - 1.16 = 2.84 \), so the y - intercept is less than 4.

Top - left graph: y - intercept at \( y = 4 \), which is higher than our calculated value.

Top - right graph: y - intercept at \( y = 4 \) or higher, and it's increasing, which is wrong (our function is decreasing).

Bottom - left graph: y - intercept around \( y = 4 \)? No, wait, maybe my calculation is wrong. Wait, \( \log_4(5)=\frac{\ln5}{\ln4}\approx1.16 \), so \( 4 - 1.16 = 2.84 \), so y - intercept is around 3.

Bottom - left graph: when \( x = 0 \), the y - value is around 4? No, maybe I made a mistake in the direction. Wait, \( y =-\log_4(x + 5)+4 \). Let's take \( x=-4 \), then \( f(-4)=4-\log_4(1)=4 - 0 = 4 \). Ah! That's a key point. When \( x=-4 \), \( f(-4)=4 \).

Now, let's check the graphs:

- Top - left graph: at \( x=-4 \), what's the y - value? The graph has a point at \( x = 0,y = 4 \), but at \( x=-4 \), it's near \( x = 0 \), no.

- Top - right graph: at \( x=-4 \), let's see. The graph has a vertical asymptote at \( x=-5 \), and at \( x=-4 \), what's the y - value? Wait, when \( x=-4 \), \( f(-4)=4 \), so the point \( (-4,4) \) should be on the graph.

- Bottom - left graph: at \( x=-4 \), what's the y - value?

- Bottom - right graph: at \( x=-4 \), \( y \) is negative? No.

Wait, when \( x=-4 \), \( f(-4)=4-\log_4(-4 + 5)=4-\log_4(1)=4 - 0 = 4 \). So the point \( (-4,4) \) is on the graph.

Now, check the graphs:

- Top - left graph: vertical asymptote at \( x = 0 \), so at \( x=-4 \), it's far from \( x = 0 \), and the y - value at \( x=-4 \) is not 4.

- Top - right graph: at \( x=-4 \), the y - value is 4 (since it passes through \( (-4,4) \)), and it's increasing? Wait, no, our function is \( f(x)=4-\log_4(x + 5) \), the derivative \( f^\prime(x)=-\frac{1}{(x + 5)\ln4} \), which is negative for \( x>-5 \), so the function is decreasing. Wait, but the top - right graph is increasing (as \( x \) increases, \( y \) increases), which contradicts the derivative.

Wait, I think I made a mistake in the derivative. Wait, \( \frac{d}{dx}\log_b(u)=\frac{u^\prime}{u\ln b} \), so \( f^\prime(x)=-\frac{1}{(x + 5)\ln4} \), which is negative, so the function is decreasing. So the graph should be decreasing.

Now, the point \( (-4,4) \) is on the graph. Let's check the bottom - left graph: at \( x=-4 \), what's the y - value? If the graph is decreasing, and at \( x=-4 \), \( y = 4 \), and as \( x \) increases (e.g., \( x = 0 \)), \( y \) decreases (to \( 4-\log_4(5)\approx2.84 \)).

The top - right graph is increasing, so it's wrong. The bottom - left graph: let's see the asymptote. The vertical asymptote is \( x=-5 \), and at \( x=-4 \), \( y = 4 \), and as \( x \) increases, \( y \) decreases. That matches.

Wait, but the user's graphs:

Top - left: asymptote at \( x = 0 \), curve on the right of \( x = 0 \), decreasing?

Top - right: asymptote at \( x=-5 \), curve on the right of \( x=-5 \), increasing.

Bottom - left: asymptote at \( x=-5 \), curve on the right of \( x=-5 \), decreasing.

Bottom - right: asymptote at \( x=-5 \), curve on the right of \( x=-5 \), decreasing but going down as \( x ightarrow - 5^+ \).

Wait, when \( x ightarrow - 5^+ \), \( f(x)=4-\log_4(x + 5) ightarrow4-(-\infty)=\infty \), so the graph should go up as \( x ightarrow - 5^+ \). So the bottom - left graph: as \( x ightarrow - 5^+ \), does it go up? Yes, because it's a decreasing function, so as \( x \) approaches \( - 5 \) from the right (increasing \( x \) towards \( - 5 \)), \( y \) approaches \( \infty \). Wait, no: if \( x \) is approaching \( - 5 \) from the right, \( x \) is getting closer to \( - 5 \), so \( x \) is decreasing towards \( - 5 \). So as \( x \) decreases towards \( - 5 \) (from the right), \( y \) increases to \( \infty \). So the graph should be increasing as \( x \) decreases towards \( - 5 \), and decreasing as \( x \) increases away from \( - 5 \).

So the correct graph is the bottom - left graph? Wait, no, the bottom - left graph: when \( x \) increases (moves to the right), \( y \) decreases, and when \( x \) decreases (moves to the left towards \( - 5 \)), \( y \) increases. That matches the derivative (since \( f^\prime(x)<0 \), so as \( x \) increases, \( y \) decreases; as \( x \) decreases, \( y \) increases).

And the point \( (-4,4) \) is on the graph (since \( f(-4)=4 \)). So the correct graph is the bottom - left graph (the third graph, if we count top - left as first, top - right as second, bottom - left as third, bottom - right as fourth).

Answer:

The bottom - left graph (the third graph in the given set of four graphs)