How do I graph a transformed rational function?

Question

If the function is $-\frac{5}{x+3} + 2$ how would I graph that without using a table of values? Do I just use the asymptotes $x=-3$, and $y= 2$ and draw the general shape? Do I use "mapping points" like $(x-3, -5y+2)$?

Answer 1

One way to graph the function $$f(x) = -\frac{5}{x + 3} + 2$$ is to apply the series of transformations $$\frac{1}{x} \to -\frac{1}{x} \to -\frac{5}{x} \to -\frac{5}{x + 3} \to -\frac{5}{x + 3} + 2$$ We start with the graph of $y = 1/x$. The $x$- and $y$-axes are, respectively, the horizontal and vertical asymptotes of the graph.

We can obtain the graph of $y = -1/x$ from the graph of $y = 1/x$ by reflecting the graph in the $x$-axis.

We can obtain the graph of $y = -5/x$ from the graph of $y = -1/x$ by stretching the graph of $y = -1/x$ vertically by a factor of $5$.

We can obtain the graph of $y = -5/(x + 3)$ from the graph of $y = -5/x$ by shifting the graph of $y = -5/x$ to the left by three units. The vertical line $x = -3$ that is shown in red is the translated vertical asymptote.

Finally, we can obtain the graph of the function $$f(x) = -\frac{5}{x + 3} + 2$$ from the graph of $y = -5/(x + 3)$ by translating the graph of $y = -5/(x + 3)$ upwards by two units. The horizontal line $y = 2$ is the translated horizontal asymptote. The vertical line $x = 3$ is the vertical asymptote.

I opted to use a larger graph for the final function in order to illustrate its asymptotic behavior.

Alternate Method: We solve for the intercepts and asymptotes, then perform a line analysis.

$x$-intercept: Setting $f(x) = 0$ yields \begin{align*} -\frac{5}{x + 3} + 2 & = 0\\ 2 & = \frac{5}{x + 3}\\ 2x + 6 & = 5\\ 2x & = -1\\ x & = -\frac{1}{2} \end{align*}

$y$-intercept: Evaluating the function at $x = 0$ yields \begin{align*} f(0) & = -\frac{5}{0 + 3} + 2\\ & = -\frac{5}{3} + \frac{6}{3}\\ & = \frac{1}{3} \end{align*}

horizontal asymptote: As $|x| \to \infty$, the term $$-\frac{5}{x + 3} \to 0$$ so $$y = -\frac{5}{x + 3} + 2 \to 2$$ the horizontal asymptote is the line $y = 2$.

vertical asymptote: Since the denominator of $$f(x) = -\frac{5}{x + 3} + 2 = -\frac{5}{x + 3} + \frac{2x + 6}{x + 3} = \frac{2x + 1}{x + 3}$$ vanishes at $x = 3$ and the numerator does not, $x = -3$ is the vertical asymptote of the function's graph.

line analysis: The function can only change sign at an $x$-intercept or a vertical asymptote. Therefore, its sign is constant in the intervals $(-\infty, -3)$, $(-3, 1/2)$, and $(1/2, \infty)$. The function is equal to $0$ at the $x$-intercept $1/2$ and undefined at $x = 3$ since it has a vertical asymptote there. Expressing the function in the form $$f(x) = \frac{2x + 1}{x + 3}$$ we see that its sign is the product of the signs of the dividend $2x + 1$ and divisor $x + 3$. The divisor is negative when $x < -3$, $0$ at $x = -3$, and positive when $x > 3$. The dividend is negative when $x < -1/2$, $0$ at $x = -1/2$, and positive when $x > 1/2$. Hence, the quotient is positive when $x < 3$, undefined at $x = -3$, negative when $-3 < x < -1/2$, zero at $x = -1/2$, and positive when $x > 1/2$.

Thus, when $x < -3$, we draw the curve in the region above the horizontal asymptote $y = 2$, and when $x > -3$, we draw the curve in the region below the horizontal asymptote so that it passes through the $x$- and $y$-intercepts, as shown above.