GCSE Exponential and trig graphs Practice Questions

Free GCSE Exponential and trig graphs practice questions with full step-by-step worked solutions. Covers exponential graph, y-intercept, k^x, asymptote. Practise exam-style problems and check your method.

exponential graphy-interceptk^xasymptotefinding ksubstitution
GCSE Higher70 questionsStep-by-step solutions
Question 1
1 markeasy
The curve y=3xy = 3^x is drawn for −2≤x≤3-2 \le x \le 3. Write down the coordinates of the point where the curve crosses the yy-axis.
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Worked solution

  1. Use the y-axis

    x=0x = 0

    Every point on the y-axis has x=0x = 0, so substitute x=0x = 0 into the equation.

  2. Substitute x=0x = 0

    y=30=1y = 3^0 = 1

    Any positive number raised to the power 00 is 11.

  3. Write the coordinates

    (0, 1)(0,\ 1)

    The curve crosses the y-axis at the point (00, 11).

Answer
(0, 1)(0,\ 1)
Question 2
1 markeasy
A student says that the graph of y=kxy = k^x, where k>0k > 0, must cross the xx-axis somewhere. Which statement correctly explains why the student is wrong?
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Worked solution

  1. Say what crossing the x-axis would mean

    y=0y = 0

    A curve crosses the x-axis at a point where its y-value is zero.

  2. Test whether kxk^x can equal 00

    kx>0 for every value of xk^x > 0 \text{ for every value of } x

    A positive number raised to any power (positive, zero or negative) is always positive.

  3. Conclude

    y=0 is an asymptotey = 0 \text{ is an asymptote}

    The curve gets closer and closer to the x-axis but never reaches it, so it never crosses it.

Answer
kx>0 for all x, so y=0 is an asymptote, never a crossing pointk^x > 0 \text{ for all } x, \text{ so } y = 0 \text{ is an asymptote, never a crossing point}
Question 3
2 marksintermediate
The number of insects in a colony is modelled by N=300×1.2 tN = 300 \times 1.2^{\,t}, where tt is the number of days. Work out the number of insects after 55 days, to the nearest whole insect.
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Worked solution

  1. Interpret the model

    1.2=100%+20%1.2 = 100\% + 20\%

    The colony grows by 2020% each day.

  2. Substitute t=5t = 5

    N=300×1.25N = 300 \times 1.2^{5}

    Five days means five lots of the multiplier.

  3. Work out the power

    1.25=2.488321.2^{5} = 2.48832

    Use a calculator and keep all the digits.

  4. Multiply

    N=300×2.48832N = 300 \times 2.48832

    Apply the five-day growth factor.

  5. Work out the value

    N=746.496N = 746.496

    This is the unrounded number of insects.

  6. Round sensibly

    N≈746N \approx 746

    You cannot have part of an insect, so round to 746746.

Answer
746 insects746 \text{ insects}
Question 4
4 markshard
Solve cos⁡x∘=0.25\cos x^\circ = 0.25 in the interval 0≤x≤7200 \le x \le 720. Give your answers correct to 11 decimal place.
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Worked solution

  1. Use the inverse cosine

    x=cos⁡−1(0.25)=75.522...∘x = \cos^{-1}(0.25) = 75.522...^\circ

    Make sure the calculator is in degree mode.

  2. Round the first solution

    x=75.5x = 75.5

    Correct to 11 decimal place.

  3. Use symmetry about x=180x = 180

    cos⁡(360−x)∘=cos⁡x∘\cos(360 - x)^\circ = \cos x^\circ

    The cosine curve is symmetrical about x=180x = 180.

  4. Find the second solution

    360−75.522...=284.477...360 - 75.522... = 284.477...

    So x=284.5x = 284.5 to 11 decimal place.

  5. Count the solutions in the first cycle

    2 solutions in 0≤x≤360\text{2 solutions in } 0 \le x \le 360

    The line y=0.25y = 0.25 cuts the cosine curve twice per cycle.

  6. Recall the period

    period of cosine=360∘\text{period of cosine} = 360^\circ

    The curve repeats identically every 360360 degrees.

  7. Extend the first solution

    75.522...+360=435.522...75.522... + 360 = 435.522...

    So x=435.5x = 435.5 to 11 decimal place.

  8. Extend the second solution

    284.477...+360=644.477...284.477... + 360 = 644.477...

    So x=644.5x = 644.5 to 11 decimal place.

  9. Check for further solutions

    435.5+360=795.5>720435.5 + 360 = 795.5 > 720

    No more solutions lie in the interval.

  10. State the solutions

    x=75.5, 284.5, 435.5, 644.5x = 75.5,\ 284.5,\ 435.5,\ 644.5

    There are four solutions in the interval 00 to 720720.

Answer
x=75.5, 284.5, 435.5, 644.5x = 75.5,\ 284.5,\ 435.5,\ 644.5
Question 5
6 markschallenging
Solve 2sin⁡x∘=cos⁡60∘2\sin x^\circ = \cos 60^\circ in the interval 0≤x≤3600 \le x \le 360. Give your answers correct to 11 decimal place.
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Worked solution

  1. Recall the exact value

    cos⁡60∘=12\cos 60^\circ = \frac{1}{2}

    This is a standard exact value.

  2. Rewrite the equation

    2sin⁡x∘=122\sin x^\circ = \frac{1}{2}

    Replace cos 6060 with one half.

  3. Isolate sin x

    sin⁡x∘=14\sin x^\circ = \frac{1}{4}

    Divide both sides by 22.

  4. Write as a decimal

    sin⁡x∘=0.25\sin x^\circ = 0.25

    One quarter is 0.250.25.

  5. Use the inverse sine

    sin⁡−1(0.25)=14.477...∘\sin^{-1}(0.25) = 14.477...^\circ

    Make sure the calculator is in degree mode.

  6. Round the first solution

    x=14.5x = 14.5

    Correct to 11 decimal place.

  7. Sketch the sine curve

    y=sin⁡x∘ and y=0.25y = \sin x^\circ \text{ and } y = 0.25

    The line cuts the positive hump twice.

  8. Use the symmetry rule

    sin⁡(180−x)∘=sin⁡x∘\sin(180 - x)^\circ = \sin x^\circ

    The sine curve is symmetrical about x=90x = 90.

  9. Find the second solution

    180−14.477...=165.522...180 - 14.477... = 165.522...

    Subtract the unrounded value from 180180.

  10. Round the second solution

    x=165.5x = 165.5

    Correct to 11 decimal place.

  11. Check no other solutions

    sin⁡x∘<0 for 180<x<360\sin x^\circ < 0 \text{ for } 180 < x < 360

    The curve is below the x-axis there, so it cannot equal 0.250.25.

  12. Verify the first

    2sin⁡14.5∘=2×0.2504=0.5008≈0.52\sin 14.5^\circ = 2 \times 0.2504 = 0.5008 \approx 0.5

    Close to cos 6060, allowing for rounding.

  13. Verify the second

    2sin⁡165.5∘=2×0.2504=0.5008≈0.52\sin 165.5^\circ = 2 \times 0.2504 = 0.5008 \approx 0.5

    Also correct.

  14. Check the interval

    0≤14.5≤360, 0≤165.5≤3600 \le 14.5 \le 360,\ 0 \le 165.5 \le 360

    Both lie in the required interval.

  15. State the solutions

    x=14.5 or x=165.5x = 14.5 \text{ or } x = 165.5

    There are exactly two solutions.

Answer
x=14.5 or x=165.5x = 14.5 \text{ or } x = 165.5

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