Curve Sketching#

Drawing the graph of \(y=f(x)\)

  • Determine the domain.

  • Find the \(x\) and \(y\) intercepts, if any.

  • Find the vertical and horizontal asymptotes, if any.

  • Determine the intervals of increase and decrease.

  • Find the relative extrema, if any.

  • Determine the intervals of concavity.

  • Find the inflection points, if any.

  • Graph some additional points to help identify the shape of \(f\), if needed, and use this information to sketch the curve.

Example 1#

Sketch the graph of

\[y = \dfrac{2x + 3}{x + 4}.\]
Step 1:   Determine the domain.

In this case, \(f(x)\) is undefined when the denominator equals zero, which is when \(x= -4\). Therefore, the domain of \(f\) is \((-\infty, -4) \cup (-4, \infty)\).

Step 2:   Find the   \(x\)   and   \(y\)   intercepts, if any.

  • \(x\)-intercept: Set \(y=0\) and solve for \(x\).

\[\begin{align*} \frac{2x + 3}{x + 4} &= 0 && \text{Set $f(x) = 0$} \\ \frac{2x + 3}{x + 4}(x + 4) &= 0(x + 4) && \text{Multiply both sides by $(x + 4)$} \\ 2x + 3 &= 0 && \text{Simplify both sides} \\ x &= -\frac{3}{2}. && \text{Solve for $x$} \end{align*}\]

  • \(y\)-intercept: Set \(x=0\) and solve for \(y\).

\[\begin{align*} y &= f(0) && \text{Plug in $x = 0$}\\ &= \frac{2(0) + 3}{0 + 4} \\ &= \frac{3}{4} \end{align*}\]

Therefore, \(f\) has an \(x\)-intercept at the point \((-3/2,0)\) and a \(y\)-intercept at the point \((0,3/4)\).

Step 3:   Find the vertical and horizontal asymptotes, if any.

  • Vertical Asymptotes: Since the numerator and denominator are already factored, the vertical asymptotes correspond to the values of \(x\) that make the denominator equal to zero. This happens only when \(x = -4\), so \(x = -4\) is the only vertical asymptote.

  • Horizontal Asymptotes: Since the degree of the numerator and the denominator are the same, both limits at infinity are equal to the ratio of leading coefficients.

    \[\lim_{x\to -\infty} \frac{2x + 3}{x + 4} = 2 = \lim_{x\to \infty} \frac{2x + 3}{x + 4}\]

    Therefore, \(y=2\) is the only horizontal asymptote.

Step 4:   Determine the intervals of increase and decrease.

  • Compute \(f'(x)\).

    \[\begin{align*} f'(x) &= \frac{2(x + 4) - (2x + 3)(1)}{(x + 4)^2} && \text{Using the quotient rule}\\ &= \frac{2x + 8 - 2x - 3}{(x + 4)^2} && \text{Simplify}\\ &= \frac{5}{(x + 4)^2} && \text{Simplify} \end{align*}\]

    Notice that \(f'(x)\) is never equal to zero and is defined for all values of \(x\) in the domain of \(f\) (i.e., \(f\) doesn’t have any critical points).

  • Compute the sign of \(f'(x)\) on the domain of \(f\).

Sign analysis of $f'(x)$

Therefore, \(f\) is increasing on \((-\infty, -4)\) and \((-4, \infty)\).

Long Text Description

A number line with positive and negative signs assigned to intervals, with positive to the left of negative four, and with positive to the right of negative four.

Step 5:   Find the relative extrema, if any.

Since \(f\) does not have any critical points, \(f\) does not have any relative extrema.

Step 6:   Determine the intervals of concavity.

  • Compute \(f''(x)\).

\[\begin{align*} f''(x) &= \frac{d}{dx} 5(x+4)^{-2} && \text{Rewrite $f'(x)$ } \\ &= -10(x+4)^{-3} \\ &= -\frac{10}{(x+4)^3} \end{align*}\]

Notice that \(f''(x)\) is never equal to zero and is defined for all values of \(x\) in the domain of \(f\).\

  • Compute the sign of \(f''(x)\) on the domain of \(f\).

Sign analysis of $f''(x)$

Therefore, \(f\) is concave up on \((-\infty, -4)\) and concave down on \((-4, \infty)\).

Long Text Description

A number line with positive and negative signs assigned to intervals, with positive to the left of negative four, and negative to the right of negative four.

Step 7:   Find the inflection points, if any.

The only change in sign of \(f''(x)\) occurs at \(x=-4\). But since \(x=-4\) is not in the domain of \(f\), \(f\) does not have any inflection points.

Step 8:   Use the information gathered to sketch the function.
Graph of $f(x)$
Long Text Description

There is a horizontal x axis with the points -16, -12, -8, -4, 4, and 8 labeled. There is a vertical y axis with the points -6, -2, 2, and 6 labeled. A curve is plotted. The curve has a horizontal asymptote represented by a blue dotted line at y=2, and a vertical asymptote represented by a red line at x= -4. The curve comes in increasing and concave up from its horizontal asymptote on the left, then heads towards infinity as x increases to -4. On the other side of x=-4, the curve comes increasing and concave down from negative infinity and increases towards its horizontal asymptote at y=2 as it heads off to the right.