Question 18
The graph of $y=f(x)$, where $f(x)$ is the quadratic function $f(x)=ax^2+bx+4$, is shown.
Two regions of the area between the graph of $y=f(x)$ and the $x$-axis are shaded.
Region $P$ has an area of $\frac{13}{6}$ units2 and Region $Q$ has an area of $\frac{43}{6}$ units2.
Determine the values of $a$ and $b$.
[4 marks]Two areas, two equations, two unknowns
Watch each shaded region become an integral with $a$ and $b$ still in it. Two areas give two equations, and elimination does the rest.
Each area is one equation in $a$ and $b$. Subtracting them removes $a$, and substituting back finds it.
See what $a$ and $b$ each do to the two areas
Move $a$ and $b$ and watch both shaded areas change. Look at the last two rows. The gap between the areas depends on $b$ alone, and their total depends on $a$ alone.
Both areas match the question, so $a=2$ and $b=5$. Subtracting the two equations is the same move as taking $Q-P$, and adding them is $P+Q$.
Part of the curve is below the $x$-axis between $-1$ and $1$. An integral counts that part as negative, so it no longer equals the shaded area. In the question the curve stays above the axis, which is why each area is just an integral.
$Q-P$ is always exactly $b$, because the $a$ and 4 parts of the two integrals are equal. The question’s gap is $\frac{43}{6}-\frac{13}{6}=5$, so set $b=5$.
The gap is right. Now the total $P+Q=\frac{2a}{3}+8$ has to equal $\frac{13}{6}+\frac{43}{6}=\frac{28}{3}$, and only $a$ changes the total.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The QCAA’s sample response writes both integrals, turns each into a linear equation, then solves by elimination.
Both shaded regions are above the $x$-axis, so each area is a definite integral of $f(x)$. Read the limits off the graph.
$$\begin{aligned}\int_{-1}^{0}f(x)\,dx&=\frac{13}{6}\\[4pt]\int_{0}^{1}f(x)\,dx&=\frac{43}{6}\end{aligned}$$
This mark may be implied by your later working.
Marker: correctly identifies the use of integrals
Integrate with $a$ and $b$ left in as letters. The constant $c$ cancels in a definite integral, so you can leave it out. Substitute the limits, then multiply through by 6 to clear the fractions.
$$\begin{aligned}\left[\frac{ax^3}{3}+\frac{bx^2}{2}+4x\right]_{-1}^{0}&=\frac{13}{6}\\0-\left(\frac{-a}{3}+\frac{b}{2}-4\right)&=\frac{13}{6}\\\frac{a}{3}-\frac{b}{2}+4&=\frac{13}{6}\\2a-3b+24&=13\\2a-3b&=-11\ \text{(A)}\end{aligned}$$
$$\begin{aligned}\left[\frac{ax^3}{3}+\frac{bx^2}{2}+4x\right]_{0}^{1}&=\frac{43}{6}\\\frac{a}{3}+\frac{b}{2}+4-(0)&=\frac{43}{6}\\2a+3b+24&=43\\2a+3b&=19\ \text{(B)}\end{aligned}$$
Equivalent forms of the two equations are accepted, such as $\frac{a}{3}-\frac{b}{2}=\frac{13}{6}-4$ and $\frac{a}{3}+\frac{b}{2}+4-(0)=\frac{43}{6}$.
Marker: correctly determines the two equations in the two unknowns $a$ and $b$
The $a$ terms match, so subtract the equations to eliminate $a$.
$$\begin{aligned}\text{(A)}-\text{(B)}:\quad-6b&=-30\\b&=5\end{aligned}$$
Follow-through marks are allowed if an error earlier changed your equations.
Marker: determines $b$
Substitute $b=5$ into (A) to find $a$.
$$\begin{aligned}2a-3(5)&=-11\\2a&=4\\a&=2\end{aligned}$$
So $f(x)=2x^2+5x+4$. A quick check against the graph: $f(-1)=1$, $f(0)=4$ and $f(1)=11$, all above the axis, with the curve rising to the right.
Marker: determines $a$
Each phrase of the question gave you something
A simple familiar area question gives you the function and asks for the area. This one runs the other way. You integrate with two unknowns still in the function, set each result equal to an area you are given, and then carry the two equations into a second procedure, solving them simultaneously by hand. It stays familiar because the graph marks the limits for you and the question names both areas, so the integrals almost write themselves.
If I had to guess where you might lose marks, it would be the algebra after the integration. The 2021 subject report says many students could not correctly manipulate these equations, which have fractional and negative coefficients, and that the successful students mostly used elimination. Watch the lower limit in region $P$. $(-1)^3=-1$ makes the $a$ term negative, and subtracting the whole bracket flips every sign inside it. Multiply both equations by 6 before you eliminate, so you are working with whole numbers.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2021 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it