Can we sketch the graph of y = (4x-1)/(x+5)?

Can we sketch the graph of $y = (4x-1)/(x+5)$ ?

Question

Sketch the graph of $y = \frac{4x-1}{x+5}.$ Hence, or otherwise, find the set of values of $x$ for which $-1 < \frac{4x-1}{x+5} < 1.$

Find the set of values of $x$ for which $3x^3 + 2x^2 + 6 \ge 19x.$

Solution

In sketching the graph, we’ll look at four different features of the function:

its intercepts with the axes;

its asymptotic behaviour;

its signs in various regions; and

its turning points, if it has any.

Intercepts

The $y$ -intercept is $y = y(0) = \frac{4 \times 0 - 1}{0+5} = -\frac{1}{5}.$

The $x$ -intercepts are the values of $x$ such that $0 = \frac{4x-1}{x+5};$ there’s one such value, $x = \dfrac{1}{4}$ .

Asymptotic behaviour

There’s a single vertical asymptote when the denominator is zero, the line $x = -5$ .

If we rewrite the function as $\frac{4x-1}{x+5}=4-\frac{21}{x+5}$ it is easier to see that as $x \to \infty$ , $y \to 4$ and that as $x \to -\infty$ , $y \to 4$ so there’s a single horizontal asymptote, $y = 4$ .

Signs

The sign of $y$ can only change when $x$ moves across a vertical asymptote or an $x$ -intercept.

The $x$ -axis falls into three regions, $x < -5$ , $-5 < x < \dfrac{1}{4}$ , and $x > \dfrac{1}{4}$ , and in each of these regions $y$ has a constant sign.

We can determine the sign of $y$ in each of these regions by looking at the signs of the numerator and denominator separately.

In the first region, $4x - 1 < 0$ and $x + 5 < 0$ , so $y$ is the quotient of two negative expressions, and is therefore positive.
In the second region, $4x - 1 < 0$ and $x + 5 > 0$ , and so $y < 0$ .
In the third region, $4x - 1 > 0$ and $x + 5 > 0$ , and so $y > 0$ .

Turning points

By considering the function in the form $\frac{4x-1}{x+5}=4-\frac{21}{x+5}$ we see that it is a transformation of $y = \dfrac{1}{x}$ . Therefore there are no turning points.

The sketch shows that we need the set of $x$ -values from $a$ to $b$ , where $a$ is the solution of $-1=\dfrac{4x-1}{x+5}$ , and $b$ is the solution to $\dfrac{4x-1}{x+5}=1$ .

We have

\begin{align*} -1&=\dfrac{4x-1}{x+5}\\ &\implies -x-5=4x-1 \\ &\implies x = -\dfrac{4}{5}. \end{align*}

and also

\begin{align*} \dfrac{4x-1}{x+5}&=1 \\ &\implies 4x-1=x+5 \\ &\implies x = 2. \end{align*}

Thus $a = -\dfrac{4}{5}$ , and $b = 2$ , and the set of values for $x$ that we need is $\left\{ x \in \mathbb{R} : -\dfrac{4}{5}< x < 2 \right\}$ .

Now $3x^3 + 2x^2 + 6 \ge 19x \iff 3x^3 + 2x^2 - 19x + 6 \ge 0,$ so if we let $p(x) = 3x^3 + 2x^2 - 19x + 6$ , then we have to find the set of values of $x$ for which $p(x) \ge 0$ .

Can we factor the polynomial $p(x)$ ?

By the factor theorem, $(x-a)$ is a factor of $p(x)$ if and only if $p(a) = 0$ .

Let’s try a few different values for $x$ : every cubic equation has at least one real root, and if this is an integer, we have a chance of finding it here.

$p(0) = 6$ ;
$p(1) = 3 + 2 - 19 + 6 = -8$ ;
$p(2) = 3 \times 8 + 2 \times 4 - 19 \times 2 + 6 = 24 + 8 - 38 + 6 = 0$ .

So our first factor is

$(x-2)$

. There are thus real constants

$\alpha$

$\beta$

and

$\gamma$

such that

\begin{align*} 3x^3 + 2x^2 - 19x + 6 = p(x) &= (\alpha x^2 + \beta x + \gamma)(x-2) \\ &= \alpha x^3 - 2\alpha x^2 + \beta x^2 - 2\beta x + \gamma x - 2\gamma \\ &= \alpha x^3 + (\beta - 2\alpha) x^2 + (\gamma - 2\beta) x - 2\gamma. \end{align*}

By equating coefficients, we have that $\alpha = 3$ , $\gamma = -3$ , and $\beta - 2\alpha = 2$ , i.e. $\beta = 8$ . Thus, $p(x) = (x-2)(3x^2 + 8x - 3).$ Factorising, we have $3x^2 + 8x - 3 = (3x-1)(x+3).$ Therefore $p(x) = (x-2)(3x-1)(x+3).$

The sign of $p(x)$ can only change as $x$ passes over a root of $p(x)$ ; these roots are $x = -3$ , $x = \dfrac{1}{3}$ , and $x = 2$ .

This divides the $x$ -axis into four regions.

In the first region, with $x < -3$ , all three factors are negative, so $p(x) < 0$ .
In the second region, with $-3 < x < \dfrac{1}{3}$ , two factors are negative and one is positive, so $p(x) > 0$ .
In the third, with $\dfrac{1}{3} < x < 2$ , one factor is negative and two are positive, and so $p(x) < 0$ .
Finally, with the fourth, with $x > 2$ , all three factors are positive and therefore $p(x) > 0$ .

We therefore have that $\{ x \in \mathbb{R} : p(x) \ge 0 \} = \{ x \in \mathbb{R} : x \ge 2 \} \cup \left\{ x \in \mathbb{R} : -3 \le x \le \frac{1}{3} \right\}.$