2024 H2 Maths Paper 1 Q1 — Finding a, b and c from a Graph

The curve, livefitting the constants

Try a value of

a

1.15

The curve The curve's lowest point Asymptotes Fixed points

Finding $a$ , $b$ and $c$

4 marks

The graph of $y=\dfrac{x+1}{ax^{2}+bx+c}$ , with $a$ , $b$ , $c$ non-zero constants, has an asymptote at $x=-\frac{1}{2}$ and a turning point at $\left(-2,\,-\frac{1}{9}\right)$ . Find the values of $a$ , $b$ and $c$ .

Approach

Three unknowns need three equations, and the question hands you exactly three facts: a vertical asymptote, a point the curve passes through, and a zero gradient at that point. Turn each into an equation in $a$ , $b$ , $c$ — in that order, because the second one produces a number that makes the third almost free.

The asymptote — the denominator vanishes there

A vertical asymptote at $x=-\frac{1}{2}$ means the denominator is zero at $x=-\frac{1}{2}$ (and the numerator is not, since $-\frac{1}{2}+1=\frac{1}{2}\neq0$ ):

a\left(-\tfrac{1}{2}\right)^{2}+b\left(-\tfrac{1}{2}\right)+c=0\ \Longrightarrow\ \frac{a}{4}-\frac{b}{2}+c=0

Multiplying through by 4:

a-2b+4c=0\qquad\cdots\ (1)

The point on the curve

Substituting $x=-2$ , $y=-\frac{1}{9}$ :

\frac{-2+1}{4a-2b+c}=-\frac{1}{9}\ \Longrightarrow\ \frac{-1}{4a-2b+c}=-\frac{1}{9}

4a-2b+c=9\qquad\cdots\ (2)

Note — keep that 9 in view

Equation (2) says the denominator of the original function takes the value $9$ at $x=-2$ . That is not just an equation to solve later; it is a number the next step needs. Reading it that way turns the quotient rule into one line of arithmetic.

The turning point — only the numerator matters

Writing $D=ax^{2}+bx+c$ for the denominator, the quotient rule gives

\frac{dy}{dx}=\frac{D\cdot(1)-(x+1)(2ax+b)}{D^{2}}

At a turning point $\frac{dy}{dx}=0$ , and a fraction is zero exactly when its numerator is — the denominator $D^{2}$ is safely non-zero at $x=-2$ , since $-2$ is not the asymptote. So it is enough that

D-(x+1)(2ax+b)=0\quad\text{at }x=-2

At $x=-2$ the two pieces are $D=4a-2b+c=9$ from (2), and $(x+1)(2ax+b)=(-1)(-4a+b)=4a-b$ . Therefore

9-(4a-b)=0\ \Longrightarrow\ 4a-b=9\qquad\cdots\ (3)

Solving the three

From (3), $b=4a-9$ . Substituting into (1) and (2):

\begin{aligned}(1):\ \ a-2(4a-9)+4c=0&\ \Longrightarrow\ -7a+18+4c=0\ \Longrightarrow\ c=\frac{7a-18}{4}\\[4pt](2):\ \ 4a-2(4a-9)+c=9&\ \Longrightarrow\ -4a+18+c=9\ \Longrightarrow\ c=4a-9\end{aligned}

Equating the two expressions for $c$ :

\begin{aligned}\frac{7a-18}{4}&=4a-9\\7a-18&=16a-36\\18&=9a\ \Longrightarrow\ a=2\end{aligned}

and then $b=4(2)-9=-1$ , $c=4(2)-9=-1$ .

To check your answer:

The denominator becomes $2x^{2}-x-1=(2x+1)(x-1)$ , which is zero at $x=-\frac{1}{2}$ ✓. At $x=-2$ , $y=\dfrac{-1}{8+2-1}=-\dfrac{1}{9}$ ✓. And the numerator of $\frac{dy}{dx}$ there is $9-(8-(-1))=9-9=0$ ✓ — all three conditions hold.

$\therefore\ a=2$ , $b=-1$ , $c=-1$ .

Where the marks go

Differentiating the quotient and then simplifying the whole thing is the slow road — and the algebra it creates is where sign errors live. At a turning point you only ever need the numerator. The other loss is treating "asymptote at $x=-\frac{1}{2}$ " as a statement about the whole function rather than about the denominator alone.

What the finished curve looks like

not asked for

Not required by the question — but a graph you can picture is a graph you can check, and this one takes ten seconds to sketch once $a$ , $b$ and $c$ are known.

With the constants found, the function is

y=\frac{x+1}{2x^{2}-x-1}=\frac{x+1}{(2x+1)(x-1)}

so it carries two vertical asymptotes, $x=-\frac{1}{2}$ and $x=1$ ; a horizontal asymptote $y=0$ , since the denominator outgrows the numerator; and it cuts the axes at $(-1,0)$ and $(0,-1)$ . Besides the given minimum at $\left(-2,-\frac{1}{9}\right)$ there is a second stationary point, a maximum at $(0,-1)$ , which the same numerator condition finds:

D-(x+1)(4x-1)=0\ \Longrightarrow\ (2x^{2}-x-1)-(4x^{2}+3x-1)=-2x^{2}-4x=0

giving $-2x(x+2)=0$ , so $x=0$ or $x=-2$ — the point already given, and one more.

To check your answer:

The question said "a turning point at $\left(-2,-\frac{1}{9}\right)$ ", not "the turning point" — and indeed there are two. That the given one drops straight out of the general condition is the strongest confirmation that the constants are right.

What this question was really testing

Reading conditions as equations. Three unknowns, three given facts, three equations. Every A-Level "find the constants" question is this, and the only skill is translating the words one at a time.
What an asymptote actually says. Vertical asymptote at $x=k$ means the denominator is zero at $x=k$ — a statement about one part of the fraction, not about $y$ .
The numerator-only shortcut. For $y=\frac{u}{v}$ , setting $\frac{dy}{dx}=0$ needs only $u'v-uv'=0$ . Squaring the denominator and carrying it through the algebra earns nothing.
Re-using what you already have. The value $D=9$ at $x=-2$ came out of equation (2) and went straight into equation (3). Noticing that is the difference between four lines and fourteen.

Three facts about a graph, three constants to pin down

Finding $a$ , $b$ and $c$

What the finished curve looks like

What this question was really testing

This is how we mark, and how we teach.

Three facts about a graph, three constants to pin down

Finding aa, bb and cc

What the finished curve looks like

What this question was really testing

This is how we mark, and how we teach.

Finding $a$ , $b$ and $c$