All revision notes topics

2.4 Key features of graphsIB Maths: Applications and Interpretation SL: Revision notes

Section 1

Maximum, minimum and the vertex

A maximum or minimum value is the highest or lowest value of the function in its domain. A turning point is where the graph changes from increasing to decreasing or the other way round. For a quadratic the turning point is the vertex. Use your GDC's maximum or minimum tool and give both coordinates: the xx-value says when or where, and the yy-value says how big. Example: A(x)=x(60−2x)A(x)=x(60-2x) has its vertex at (15,450)(15, 450), so the largest area is 450 m2^2 when the width is 15 m. On a restricted domain the greatest or least value may be at an end point, so compare the turning points with the values at the ends.

Key termsmaximumminimumvertexturning point
Common mistake

Giving only the xx-value of a maximum. State the maximum value (the yy-value) as well, with units.

Section 2

Intercepts, zeros and roots

The yy-intercept is the point where x=0x=0. The zeros of ff are the xx-values where f(x)=0f(x)=0; they are the xx-intercepts of the graph, and the roots of the equation f(x)=0f(x)=0. Use your GDC's zero (root) tool, or the equation solver. Example: A(x)=x(60−2x)A(x)=x(60-2x) has zeros x=0x=0 and x=30x=30. In context, x=30x=30 would leave no fencing for the length, so the area is zero.

Key termszerorootintercept
Exam tip

Check that each zero is inside the domain; reject any that is not (a negative time, for example).

Section 3

Symmetry

The graph of a quadratic y=ax2+bx+cy=ax^2+bx+c is symmetric about a vertical line through its vertex, the axis of symmetry, x=−b2ax=-\frac{b}{2a}. It lies halfway between the two zeros. Example: if the zeros are x=0x=0 and x=30x=30, the axis of symmetry is x=15x=15. For A(x)=−2x2+60xA(x)=-2x^2+60x: −602(−2)=15-\frac{60}{2(-2)}=15. Symmetry lets you find the vertex from the zeros: the xx-value is the midpoint.

Key termsaxis of symmetry
Common mistake

Using −ba-\frac{b}{a} instead of −b2a-\frac{b}{2a}.

Section 4

Asymptotes

An asymptote is a line that the graph gets closer and closer to but does not reach. A vertical asymptote is where the function is undefined, e.g. n=0n=0 for C(n)=600n+15C(n)=\frac{600}{n}+15. A horizontal asymptote shows the long-term value as xx becomes large. For C(n)=600n+15C(n)=\frac{600}{n}+15: as n→∞n\to\infty, 600n→0\frac{600}{n}\to0, so the horizontal asymptote is C=15C=15. For an exponential model such as T=20+70×0.9tT=20+70\times0.9^t, the asymptote is T=20T=20. View the graph on your GDC and write the asymptote as an equation.

Key termsvertical asymptotehorizontal asymptote
Exam tip

In context, the horizontal asymptote is a limiting value: the model approaches it but never reaches it.

Section 5

Points of intersection

To find where two curves or lines meet, graph both on your GDC and use the intersection tool. The xx-coordinates satisfy f(x)=g(x)f(x)=g(x). Example: P(t)=500×1.3tP(t)=500\times1.3^t and Q(t)=2000+300tQ(t)=2000+300t meet at t=8.39t=8.39. You can also graph the difference, Q(t)−P(t)Q(t)-P(t), and find its zeros and its maximum: the greatest gap is 18001800 at t=3.15t=3.15. Give answers to 3 significant figures and interpret them in context, with units.

Key termsintersection
Common mistake

Stating only one coordinate. Give the xx-value and, if asked for the value of the function, the yy-value too.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on 2.4 Key features of graphs

  1. A farmer uses 60 m of fencing to make a rectangular pen against a straight wall, so only three sides need fencing. The width of the pen, perpendicular to the wall, is xx metres, and its area is A(x)=x(60−2x)A(x)=x(60-2x) m2^2.
    Write down the zeros of AA, and explain what the larger zero means in this context.2 marks
  2. A school hires a coach for a trip. The cost per student, CC AED, is modelled by C(n)=600n+15C(n)=\frac{600}{n}+15, where n≥1n\ge1 is the number of students on the trip.
    Explain what the horizontal asymptote means in this context.2 marks
  3. Two colonies of bacteria are grown in a laboratory. For 0≤t≤120\le t\le12 days, the number of bacteria in colony A is P(t)=500×1.3tP(t)=500\times1.3^t and the number in colony B is Q(t)=2000+300tQ(t)=2000+300t. Use your GDC where appropriate.
    (i) Write down the number of bacteria in each colony at t=0t=0. (ii) Find the value of tt, for t>0t>0, at which the two colonies have the same number of bacteria.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).