GCSE Gradient and intercepts Practice Questions

Free GCSE Gradient and intercepts practice questions with full step-by-step worked solutions. Covers reading a graph, y-intercept, gradient, rise over run. Practise exam-style problems and check your method.

reading a graphy-interceptgradientrise over runy = mx + cgradient between two points
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
The straight line shown on the graph crosses the yy-axis. Write down the coordinates of the point where the line crosses the yy-axis.
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Worked solution

  1. Find the y-axis

    x=0x = 0

    The y-axis is the vertical line where x=0x = 0.

  2. Read where the line crosses

    y=2 when x=0y = 2 \text{ when } x = 0

    Follow the line to where it cuts the y-axis; it passes through a height of 2.

  3. Write the coordinates

    (0,2)(0, 2)

    The crossing point is written as a coordinate: x=0x = 0, y=2y = 2.

Answer
(0,2)(0, 2)
Question 2
2 markseasy
A straight line has equation y=2x+5y = 2x + 5. Write down the coordinates of the point where it crosses the yy-axis.
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Worked solution

  1. On the y-axis, x=0x = 0

    x=0x = 0

    Any point on the y-axis has x-coordinate 0.

  2. Substitute x=0x = 0

    y=2(0)+5=5y = 2(0) + 5 = 5

    The constant term gives the height on the y-axis.

  3. State the point

    (0,5)(0, 5)

    The line crosses the y-axis at (0, 5).

Answer
(0,5)(0, 5)
Question 3
2 marksintermediate
A line has equation 2y=4x+62y = 4x + 6. Work out its gradient.
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Worked solution

  1. The equation is not yet y = ...

    2y=4x+62y = 4x + 6

    We must make y the subject to read the gradient.

  2. Divide every term by 2

    2y2=4x2+62\frac{2y}{2} = \frac{4x}{2} + \frac{6}{2}

    Divide both sides by 2.

  3. Simplify

    y=2x+3y = 2x + 3

    Now it is in the form y = mx + c.

  4. Identify the gradient

    m=2m = 2

    The coefficient of x is 2.

  5. Note the intercept

    c=3c = 3

    The y-intercept is 3, though not asked for.

  6. State the gradient

    m=2m = 2

    The gradient of the line is 2.

Answer
22
Question 4
4 markshard
Line P has equation y=3x+2y = 3x + 2 and line Q has equation 2y=10x−42y = 10x - 4. Work out the gradient of the steeper line.
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Worked solution

  1. Read the gradient of P

    y=3x+2⇒mP=3y = 3x + 2 \Rightarrow m_P = 3

    Line P is already in y = mx + c form.

  2. Line Q is not in that form

    2y=10x−42y = 10x - 4

    We must make y the subject.

  3. Divide every term by 2

    2y2=10x2−42\frac{2y}{2} = \frac{10x}{2} - \frac{4}{2}

    Divide both sides by 2.

  4. Simplify

    y=5x−2y = 5x - 2

    Now Q is in y = mx + c form.

  5. Read the gradient of Q

    mQ=5m_Q = 5

    The coefficient of x is 5.

  6. Compare the two gradients

    3 and 53 \text{ and } 5

    Both are positive, so compare their sizes.

  7. Choose the larger

    5>35 > 3

    Five is larger than three.

  8. Identify the steeper line

    line Q\text{line Q}

    Line Q rises more steeply.

  9. State its gradient

    55

    The steeper line has gradient 5.

  10. Sense check

    Q rises 5 per 1\text{Q rises } 5 \text{ per } 1

    Q climbs faster than P, confirming it is steeper.

Answer
55
Question 5
6 markschallenging
A hot drink cools so that its temperature is T=80−4tT = 80 - 4t degrees Celsius after tt minutes. Work out how many minutes it takes for the drink to cool to 2020 degrees Celsius.
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Worked solution

  1. Read the formula

    T=80−4tT = 80 - 4t

    The drink starts at 80°C and cools as time passes.

  2. Interpret the intercept

    80=start temperature80 = \text{start temperature}

    At t=0t = 0 the drink is 80∘80^\circC.

  3. Interpret the gradient

    −4=°C per minute-4 = \text{°C per minute}

    The temperature falls by 4°C each minute.

  4. Note the sign

    negative⇒cooling\text{negative} \Rightarrow \text{cooling}

    A negative gradient means the temperature is dropping.

  5. Set the temperature to 20

    20=80−4t20 = 80 - 4t

    We want the time when T=20∘T = 20^\circC.

  6. Subtract 80 from both sides

    20−80=−4t20 - 80 = -4t

    Move the starting temperature across.

  7. Simplify the left side

    −60=−4t-60 = -4t

    Twenty minus eighty is minus sixty.

  8. Divide by -4

    t=−60−4t = \frac{-60}{-4}

    Dividing two negatives gives a positive.

  9. Work it out

    t=15t = 15

    Sixty divided by four is fifteen.

  10. Add units

    15 minutes15 \text{ minutes}

    It takes 15 minutes to cool to 20°C.

  11. Check the start

    80−4(0)=8080 - 4(0) = 80

    At the start the drink is 80°C.

  12. Check the answer

    80−4(15)=2080 - 4(15) = 20

    After 15 minutes the temperature is 20°C.

  13. How much it has cooled

    80−20=6080 - 20 = 60

    The drink has cooled by 60°C in total.

  14. Relate to the rate

    604=15\frac{60}{4} = 15

    Cooling 60°C at 4°C per minute takes 15 minutes.

  15. State the answer

    15 minutes15 \text{ minutes}

    The drink reaches 20°C after 15 minutes.

Answer
15 minutes

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