Review question

# Can we sketch the graph of $y = (4x-1)/(x+5)$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6317

## Solution

1. Sketch the graph of $y = \frac{4x-1}{x+5}.$

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

1. its intercepts with the axes;
2. its asymptotic behaviour;
3. its signs in various regions; and
4. 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.

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

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 $\left\{ x \in \mathbb{R} : -\dfrac{4}{5}< x < 2 \right\}$.

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

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