Review question

# Can we sketch $y=ax/(x^2+x+1)$ when $a$ is positive? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7129

## Solution

If $y=\frac{ax}{x^2+x+1} \qquad (a>0)$ show that for real values of $x$ $-a\leq y \leq \frac{1}{3}a.$

Multiplying through by the denominator gives $x^2y+(y-a)x+y = 0,$ which we can think of as a quadratic in $x$.

To ensure that $x$ is always real, we need the discriminant to be non-negative, which means \begin{align*} 0 &\le (y-a)^2-4y^2 \\ \iff 0 &\le y^2 -2ay+a^2-4y^2 \\ \iff 0 &\le -3y^2 -2ay+a^2 \\ \iff 0 &\ge 3y^2 +2ay-a^2 \\ \iff 0 &\ge (3y-a)(y+a). \end{align*}

From the sketch, we can see that this inequality is satisfied when $-a \le y \le \frac{a}{3}$ as required.

Sketch the graph of $y$ in the case $a=6$.

Firstly, we can check for vertical asymptotes occurring when the denominator is $0$. $x^2+x+1=0 \Longrightarrow x = \frac{-1\pm \sqrt{-3}}{2}.$ These roots are not real, so the denominator of $f(x)$ is never zero.

When $x=0$, we have $y=0$. This is the only intersection with the axes.

As $x \rightarrow \infty$, $y$ behaves like $\dfrac{1}{x}$ and so $y \rightarrow 0$.

Looking for stationary points, we can differentiate using the quotient rule and set $\dfrac{dy}{dx}=0$: \begin{align*} \frac{dy}{dx} = \frac{6(x^2+x+1)-6x(2x+1)}{(x^2+x+1)^2} &= 0\\ \iff \qquad -6x^2+6 &=0,\\ \end{align*}

and this holds if and only if $x=\pm 1$.

When $x = 1$, we have $y = 2$, and when $x = -1$ we have $y = -6$ (which are precisely $\dfrac{a}{3}$ and $-a$ as expected).

By drawing on the same diagram a certain straight line show that the equation $x^3-1=6x$ has three real roots, two negative and one positive.

Note that $x^3-1=(x-1)(x^2+x+1)$ so the equation $x^3-1=6x$ is equivalent to $x-1= \frac{6x}{x^2+x+1}$ (we saw above that $x^2 + x + 1 \neq 0$ for real $x$).

Thus, adding the line $y=x-1$,

we can see that $x^3 - 1 = 6x$ has two negative roots and one positive.