Question

graph the following rational function.
$f(x) = \frac{-x - 4}{x^2 - 36}$
step 3 of 3 : sketch the graph of the function.
answer
the curve shape for each region can be drawn by first selecting the region then choosing the desired shape.
the curve shape for each region can be drawn by first selecting the region then choosing the desired shape.

Explanation:

Step1: Analyze the function

The function is \( f(x)=\frac{-x - 4}{x^{2}-36}=\frac{-(x + 4)}{(x - 6)(x + 6)} \). The vertical asymptotes are at \( x=-6 \) and \( x = 6 \), and the horizontal asymptote is \( y = 0 \) (since the degree of the numerator is less than the degree of the denominator).

Step2: Determine intervals

The intervals to consider are \( (-\infty,-6) \), \( (-6,6) \), and \( (6,\infty) \).

For \( (-\infty,-6) \):

Pick a test point, say \( x=-7 \). Then \( f(-7)=\frac{-(-7)-4}{(-7)^{2}-36}=\frac{7 - 4}{49 - 36}=\frac{3}{13}\approx0.23>0 \). As \( x ightarrow-\infty \), \( f(x)\approx\frac{-x}{x^{2}}=\frac{-1}{x} ightarrow0^{+} \) (since \( x \) is negative, \( \frac{-1}{x}>0 \)). As \( x ightarrow - 6^{-} \), the denominator \( (x - 6)(x + 6) ightarrow(-12)(0^{-}) = 0^{+} \), and the numerator \( -x - 4 ightarrow -(-6)-4 = 2>0 \), so \( f(x) ightarrow+\infty \). So the shape in \( (-\infty,-6) \) should be a curve that approaches 0 from above as \( x ightarrow-\infty \) and goes to \( +\infty \) as \( x ightarrow - 6^{-} \), which matches the top - right shape (second shape in the first row).

For \( (-6,6) \):

Pick a test point, say \( x = 0 \). Then \( f(0)=\frac{-0 - 4}{0 - 36}=\frac{-4}{-36}=\frac{1}{9}\approx0.11>0 \). Pick \( x = 5 \), \( f(5)=\frac{-5 - 4}{25 - 36}=\frac{-9}{-11}=\frac{9}{11}\approx0.82>0 \). Pick \( x=-5 \), \( f(-5)=\frac{-(-5)-4}{25 - 36}=\frac{5 - 4}{-11}=\frac{1}{-11}\approx - 0.09<0 \). Wait, we made a mistake in the test point. Let's take \( x=-5 \): \( f(-5)=\frac{-(-5)-4}{(-5)^{2}-36}=\frac{5 - 4}{25 - 36}=\frac{1}{-11}<0 \). As \( x ightarrow - 6^{+} \), denominator \( (x - 6)(x + 6) ightarrow(-12)(0^{+})=0^{-} \), numerator \( -x - 4 ightarrow -(-6)-4 = 2>0 \), so \( f(x) ightarrow-\infty \). As \( x ightarrow6^{-} \), denominator \( (x - 6)(x + 6) ightarrow(0^{-})(12)=0^{-} \), numerator \( -x - 4 ightarrow - 6 - 4=-10<0 \), so \( f(x) ightarrow+\infty \) (since negative divided by negative is positive). At \( x = 0 \), \( f(0)=\frac{1}{9}>0 \). So the function in \( (-6,6) \): when \( x\in(-6,-4) \) (root of numerator is \( x=-4 \)), numerator \( -x - 4>0 \) when \( x < - 4 \), and \( <0 \) when \( x>-4 \). So for \( x\in(-6,-4) \), numerator positive, denominator negative (since \( x^{2}-36<0 \) in \( (-6,6) \)), so \( f(x)<0 \). For \( x\in(-4,6) \), numerator negative, denominator negative, so \( f(x)>0 \). As \( x ightarrow - 6^{+} \), \( f(x) ightarrow-\infty \); as \( x ightarrow6^{-} \), \( f(x) ightarrow+\infty \). The shape in \( (-6,6) \): from \( x=-6^{+} \) (going from \( -\infty \)) to \( x=-4 \) (where \( f(-4)=\frac{-(-4)-4}{16 - 36}=\frac{0}{-20}=0 \)), then from \( x=-4 \) to \( x = 6^{-} \) (going to \( +\infty \)). The middle - left shape (second shape in the second row) or the middle - right? Wait, when \( x\in(-6,-4) \), \( f(x)<0 \), and when \( x\in(-4,6) \), \( f(x)>0 \). At \( x=-4 \), \( f(x) = 0 \). As \( x ightarrow - 6^{+} \), \( f(x) ightarrow-\infty \), and as \( x ightarrow - 4^{-} \) (in \( (-6,-4) \)), \( f(x) \) goes from \( -\infty \) to 0 (at \( x=-4 \)). Then from \( x=-4 \) to \( x = 6^{-} \), \( f(x) \) goes from 0 to \( +\infty \). The shape that goes from \( -\infty \) to 0 and then to \( +\infty \) is the S - shaped curve? No, wait, let's re - evaluate. The numerator is \( -x - 4=-(x + 4) \), so the x - intercept is at \( x=-4 \). The denominator is \( (x - 6)(x + 6) \), vertical asymptotes at \( x=\pm6 \). In \( (-6,6) \), the function has a root at \( x=-4 \). For \( x\in(-6,-4) \), \( f(x)=\frac{-(x + 4)}{(x - 6)(x + 6)} \), \( x + 4<0 \), \( x - 6<0 \), \( x + 6>0 \), so denominator \( (x - 6)(x + 6)<0 \), numerator \( -(x + 4)>0 \) (since \( x + 4<0 \)), so \( f(x)<0 \). For \( x\in(-4,6) \), \( x + 4>0 \), numerator \( -(x + 4)<0 \), denominator \( (x - 6)(x + 6)<0 \), so \( f(x)>0 \). So the graph in \( (-6,6) \) comes from \( -\infty \) (as \( x ightarrow - 6^{+} \)), crosses the x - axis at \( x=-4 \), and goes to \( +\infty \) (as \( x ightarrow6^{-} \)). The shape that matches this is the third - row left shape (the S - like curve? No, the third - row left shape is a curve that goes from top to bottom? Wait, no. Wait, when \( x\in(-6,-4) \), \( f(x)<0 \), so the graph is below the x - axis, and when \( x\in(-4,6) \), \( f(x)>0 \), above the x - axis. So from \( x=-6^{+} \) (below x - axis, going to \( -\infty \)) to \( x=-4 \) (crossing x - axis), then from \( x=-4 \) (above x - axis) to \( x = 6^{-} \) (going to \( +\infty \)). The shape that has a part below and a part above, crossing the axis, is the third - row left shape (the curve that goes from the top - left to bottom - right in its square? No, the third - row left shape is a curve that is decreasing then increasing? Wait, maybe we made a mistake. Let's use the test point \( x = 5 \): \( f(5)=\frac{-5 - 4}{25 - 36}=\frac{-9}{-11}=\frac{9}{11}\approx0.82>0 \), \( x = - 5 \): \( f(-5)=\frac{-(-5)-4}{25 - 36}=\frac{1}{-11}\approx - 0.09<0 \), \( x = 0 \): \( f(0)=\frac{-4}{-36}=\frac{1}{9}\approx0.11>0 \). So the graph in \( (-6,6) \): at \( x=-6^{+} \), \( f(x) ightarrow-\infty \); at \( x=-4 \), \( f(x)=0 \); at \( x = 0 \), \( f(x)>0 \); at \( x = 6^{-} \), \( f(x) ightarrow+\infty \). So the shape is a curve that comes from \( -\infty \) (left of \( x=-6 \) no, in \( (-6,6) \)), crosses the x - axis at \( x=-4 \), and goes to \( +\infty \) as \( x ightarrow6^{-} \). The shape that matches this is the second shape in the third row? No, the third - row left shape is a curve that has a "s" shape but with a negative part and a positive part. Wait, maybe the correct shape for \( (-6,6) \) is the third - row left shape (the curve that goes from the upper left to lower right in its square? No, let's think again.

For \( (6,\infty) \):

Pick a test point, say \( x = 7 \). Then \( f(7)=\frac{-7 - 4}{49 - 36}=\frac{-11}{13}\approx - 0.85<0 \). As \( x ightarrow+\infty \), \( f(x)\approx\frac{-x}{x^{2}}=\frac{-1}{x} ightarrow0^{-} \) (since \( x \) is positive, \( \frac{-1}{x}<0 \)). As \( x ightarrow6^{+} \), the denominator \( (x - 6)(x + 6) ightarrow(0^{+})(12)=0^{+} \), and the numerator \( -x - 4 ightarrow - 6 - 4=-10<0 \), so \( f(x) ightarrow-\infty \). So the shape in \( (6,\infty) \) should be a curve that approaches 0 from below as \( x ightarrow+\infty \) and goes to \( -\infty \) as \( x ightarrow6^{+} \), which matches the top - left shape (first shape in the first row) but mirrored? Wait, no. As \( x ightarrow+\infty \), \( f(x) ightarrow0^{-} \), and as \( x ightarrow6^{+} \), \( f(x) ightarrow-\infty \), so the curve comes from \( -\infty \) (as \( x ightarrow6^{+} \)) and approaches 0 from below as \( x ightarrow+\infty \), which matches the first shape in the first row (the curve that goes from \( -\infty \) to 0 from below).

To sketch the graph:
- For \( (-\infty,-6) \): Use the second shape in the first row (top - right curve).
- For \( (-6,6) \): Use the third - row left curve (the S - like curve that crosses the x - axis at \( x=-4 \)).
- For \( (6,\infty) \): Use the first shape in the first row (top - left curve).

Answer:

To sketch the graph of \( f(x)=\frac{-x - 4}{x^{2}-36} \):

• In the interval \( (-\infty,-6) \), use the second shape in the first row (the curve that approaches \( 0^{+} \) as \( x

ightarrow-\infty \) and \( +\infty \) as \( x
ightarrow - 6^{-} \)).

• In the interval \( (-6,6) \), use the third - row left shape (the curve that crosses the x - axis at \( x = - 4 \), comes from \( -\infty \) as \( x

ightarrow - 6^{+} \) and goes to \( +\infty \) as \( x
ightarrow6^{-} \)).

• In the interval \( (6,\infty) \), use the first shape in the first row (the curve that approaches \( 0^{-} \) as \( x

ightarrow+\infty \) and \( -\infty \) as \( x
ightarrow6^{+} \)).

(Note: Since the problem asks to sketch the graph by selecting regions and shapes, the key is to identify the correct shape for each interval as described above. The final graph is constructed by combining these shapes in their respective intervals.)