Question

graph the following rational function.
$f(x) = \frac{-x + 1}{x^2 - 64}$
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.

Explanation:

Step1: Analyze Vertical Asymptotes

The denominator is \(x^2 - 64=(x - 8)(x + 8)\), so vertical asymptotes at \(x = 8\) and \(x=-8\). The domain is split into \(x < - 8\), \(-8 < x < 8\), and \(x>8\).

Step2: Analyze Horizontal Asymptote

Degree of numerator (\(1\)) < degree of denominator (\(2\)), so horizontal asymptote \(y = 0\) (the x - axis).

Step3: Analyze Sign and Behavior in Regions

- For \(x < - 8\) (e.g., \(x=-9\)): \(f(-9)=\frac{-(-9)+1}{(-9)^2 - 64}=\frac{10}{17}>0\). As \(x ightarrow-\infty\), \(f(x) ightarrow0^+\); as \(x ightarrow - 8^-\), \(f(x) ightarrow+\infty\) (since denominator \( ightarrow0^+\) and numerator \( ightarrow -(-8)+1 = 9>0\)). So the curve in \(x < - 8\) is in the second quadrant, increasing towards \(x=-8^-\) and approaching \(y = 0\) as \(x ightarrow-\infty\) (matches the top - left curve shape? Wait, no. Wait, when \(x < - 8\), let's take \(x=-10\): \(f(-10)=\frac{-(-10)+1}{100 - 64}=\frac{11}{36}>0\). So the function is positive here. As \(x\) approaches \(-8\) from the left (\(x ightarrow - 8^-\)), denominator \((x - 8)(x + 8)\) approaches \(0\) from the positive side (since \(x+8 ightarrow0^-\), \(x - 8 ightarrow - 16\), so \((x - 8)(x + 8) ightarrow0^-\)? Wait, no: \(x=-9\), \(x + 8=-1\), \(x - 8=-17\), product \(=17>0\). \(x=-7.9\) (close to \(-8\) from left), \(x + 8=-0.1\), \(x - 8=-15.9\), product \(=1.59>0\). Wait, \(x^2-64=(x - 8)(x + 8)\), when \(x < - 8\), \(x-8<0\), \(x + 8<0\), so product \(>0\). When \(-8 < x < 8\), \(x - 8<0\), \(x + 8>0\), product \(<0\). When \(x>8\), \(x - 8>0\), \(x + 8>0\), product \(>0\).

- For \(-8 < x < 8\) (e.g., \(x = 0\)): \(f(0)=\frac{1}{-64}<0\). As \(x ightarrow - 8^+\), denominator \( ightarrow0^-\), numerator \( ightarrow -(-8)+1 = 9>0\), so \(f(x) ightarrow-\infty\). As \(x ightarrow8^-\), denominator \( ightarrow0^-\), numerator \( ightarrow - 8 + 1=-7<0\), so \(f(x) ightarrow+\infty\). At \(x = 1\), \(f(1)=\frac{-1 + 1}{1 - 64}=0\) (x - intercept at \(x = 1\)). So in \(-8 < x < 8\), the function is negative except at \(x = 1\) (where it crosses the x - axis). The shape here: between \(x=-8\) and \(x = 1\), as \(x\) increases from \(-8^+\) to \(1\), \(f(x)\) goes from \(-\infty\) to \(0\) (decreasing? Wait, take \(x=-7\): \(f(-7)=\frac{-(-7)+1}{49 - 64}=\frac{8}{-15}<0\). \(x = 0\): \(f(0)=-\frac{1}{64}<0\). \(x = 1\): \(f(1)=0\). So from \(x=-8^+\) to \(x = 1\), the function is negative, increasing from \(-\infty\) to \(0\). From \(x = 1\) to \(x = 8^-\), \(f(x)\) is negative (wait, \(x = 2\): \(f(2)=\frac{-2 + 1}{4 - 64}=\frac{-1}{-60}=\frac{1}{60}>0\)? Wait, no: \(f(2)=\frac{-2 + 1}{4-64}=\frac{-1}{-60}=\frac{1}{60}>0\). Oh, I made a mistake. At \(x = 2\), numerator is \(-2 + 1=-1\), denominator is \(4 - 64=-60\), so \(\frac{-1}{-60}=\frac{1}{60}>0\). So in \(-8 < x < 8\), for \(x>1\), numerator \(-x + 1<0\) when \(x>1\), denominator \(x^2-64<0\) when \(-8 < x < 8\), so negative divided by negative is positive. For \(x < 1\) (and \(-8 < x < 8\)), numerator \(-x + 1>0\) (when \(x < 1\)), denominator \(x^2-64<0\), so positive divided by negative is negative. So the function crosses the x - axis at \(x = 1\). In \(-8 < x < 1\): negative, as \(x ightarrow - 8^+\), \(f(x) ightarrow-\infty\); as \(x ightarrow1^-\), \(f(x) ightarrow0^-\). In \(1 < x < 8\): positive, as \(x ightarrow1^+\), \(f(x) ightarrow0^+\); as \(x ightarrow8^-\), \(f(x) ightarrow+\infty\). So the shape in \(-8 < x < 8\) has a "S" - like shape? Wait, no, between two vertical asymptotes, with a zero at \(x = 1\). The middle region (\(-8 < x < 8\)): left part (\(-8 < x < 1\)) is negative, right part (\(1 < x < 8\)) is positive, crossing at \(x = 1\). So the shape here is like the middle - left or middle - right? Wait, the middle row of curve shapes: the left - middle (third row first) is a curve that goes from bottom to top (negative to positive?) No, the third row first: it's a curve that is in the left, going from top to bottom? Wait, maybe better to use the horizontal asymptote \(y = 0\) and vertical asymptotes.

- For \(x>8\) (e.g., \(x = 9\)): \(f(9)=\frac{-9 + 1}{81 - 64}=\frac{-8}{17}<0\). As \(x ightarrow+\infty\), \(f(x) ightarrow0^-\); as \(x ightarrow8^+\), denominator \( ightarrow0^+\) (since \(x - 8 ightarrow0^+\), \(x + 8 ightarrow16\)), numerator \( ightarrow - 8+1=-7<0\), so \(f(x) ightarrow-\infty\) as \(x ightarrow8^+\). So in \(x>8\), the function is negative, as \(x ightarrow8^+\), \(f(x) ightarrow-\infty\), and as \(x ightarrow+\infty\), \(f(x) ightarrow0^-\). So the shape here is like the top - right? No, the bottom - right? Wait, the curve shapes:

Now, to sketch:

1. Region \(x < - 8\): \(f(x)>0\), horizontal asymptote \(y = 0\) (approaching from above as \(x ightarrow-\infty\)), vertical asymptote \(x=-8\) (approaching \(+\infty\) as \(x ightarrow - 8^-\)). So the curve is in the second quadrant, increasing towards \(x=-8^-\) and approaching \(y = 0\) as \(x ightarrow-\infty\) (matches the top - left curve (first row first) which is a curve in the left, going from near \(y = 0\) up to \(+\infty\) as \(x ightarrow - 8^-\)).

2. Region \(-8 < x < 8\): crosses the x - axis at \(x = 1\). Left of \(x = 1\) (\(-8 < x < 1\)): \(f(x)<0\), so below the x - axis, approaching \(-\infty\) as \(x ightarrow - 8^+\) and approaching \(0\) as \(x ightarrow1^-\). Right of \(x = 1\) (\(1 < x < 8\)): \(f(x)>0\), above the x - axis, approaching \(+\infty\) as \(x ightarrow8^-\) and approaching \(0\) as \(x ightarrow1^+\). So the shape here is like the middle - left curve (third row first) but flipped? Wait, the third row second (middle - right) is a curve that goes from below to above, crossing the x - axis, which matches the behavior in \(-8 < x < 8\) (crossing at \(x = 1\), negative left, positive right).

3. Region \(x>8\): \(f(x)<0\), horizontal asymptote \(y = 0\) (approaching from below as \(x ightarrow+\infty\)), vertical asymptote \(x = 8\) (approaching \(-\infty\) as \(x ightarrow8^+\)). So the curve is in the fourth quadrant, decreasing towards \(x = 8^+\) and approaching \(y = 0\) as \(x ightarrow+\infty\) (matches the top - right curve? No, the first row second is a curve in the right, going from \(+\infty\) down to near \(y = 0\), but we need negative here. Wait, no, in \(x>8\), \(f(x)<0\), so it's below the x - axis. As \(x ightarrow8^+\), \(f(x) ightarrow-\infty\) (since numerator \(-x + 1 ightarrow - 7\), denominator \( ightarrow0^+\), so \(\frac{-7}{0^+} ightarrow-\infty\)), and as \(x ightarrow+\infty\), \(f(x) ightarrow0^-\). So the curve is in the fourth quadrant, decreasing from \(-\infty\) (as \(x ightarrow8^+\)) to \(0^-\) as \(x ightarrow+\infty\), which matches the first row second curve but flipped? Wait, maybe the correct way is to use the given curve shapes:

- For \(x < - 8\): use the first row first curve (left - top) which is a curve in the left, going from near \(y = 0\) up to \(+\infty\) (since \(f(x)>0\) here).

- For \(-8 < x < 8\): use the third row second curve (middle - right) which crosses the x - axis, has negative left and positive right.

- For \(x>8\): use the first row second curve (right - top) but flipped? No, wait, the first row second is a curve in the right, going from \(+\infty\) down to near \(y = 0\), but we need negative, so it should be a curve in the right, going from \(-\infty\) up to near \(y = 0\) (but below the x - axis). Wait, maybe the correct curve shapes are:

After analyzing the sign and asymptotes, we can sketch the graph by placing the curves in each region:

- Left of \(x=-8\): curve in the second quadrant, above x - axis, approaching \(y = 0\) as \(x ightarrow-\infty\) and \(+\infty\) as \(x ightarrow - 8^-\) (first row first curve).

- Between \(x=-8\) and \(x = 8\): curve crossing x - axis at \(x = 1\), below x - axis left of \(x = 1\), above x - axis right of \(x = 1\) (third row second curve).

- Right of \(x = 8\): curve in the fourth quadrant, below x - axis, approaching \(y = 0\) as \(x ightarrow+\infty\) and \(-\infty\) as \(x ightarrow8^+\) (first row second curve flipped? No, maybe the first row second curve is for positive, but we need negative, so maybe the fourth row second? No, the key is to use the asymptotes and sign to select the correct curve shapes for each region.

Answer:

To sketch the graph:

1. For \(x < - 8\): Use the top - left curve (first row, first column) as \(f(x)>0\), approaches \(y = 0\) as \(x

ightarrow-\infty\) and \(+\infty\) as \(x
ightarrow - 8^-\).

1. For \(-8 < x < 8\): Use the middle - right curve (third row, second column) as it crosses the x - axis, has \(f(x)<0\) for \(x < 1\) and \(f(x)>0\) for \(x>1\) in this interval.
2. For \(x>8\): Use the top - right curve (first row, second column) but since \(f(x)<0\) here, it is a reflection over the x - axis of the top - right curve (or recognize that the shape is a curve in the fourth quadrant, approaching \(y = 0\) as \(x

ightarrow+\infty\) and \(-\infty\) as \(x
ightarrow8^+\), which matches the top - right curve's shape but below the x - axis).

(Note: Since the problem is about sketching, the final answer is the correct placement of the curve shapes in each region as per the above analysis. The actual sketching involves selecting the appropriate curve from the given options for each of the three regions (\(x < - 8\), \(-8 < x < 8\), \(x>8\)) based on the function's asymptotes, sign, and intercepts.)