Graphs

1. 1

   A straight line passes through $(0, 3)$ and $(2, 7)$. What is its equation?

   • A

     $y = 2x + 3$

   • B

     $y = 2x - 3$

   • C

     $y = 4x + 3$

   • D

     $y = \frac{1}{2}x + 3$

   Check answer

   Answer & explanation

   Correct: option A

   Gradient $= (7-3)/(2-0) = 2$. The $y$-intercept is $3$ (given). So $y = 2x + 3$.

2. 2

   Which line is perpendicular to $y = 3x - 1$?

   • A

     $y = -\frac{1}{3}x + 4$

   • B

     $y = 3x + 2$

   • C

     $y = \frac{1}{3}x - 1$

   • D

     $y = -3x + 1$

   Check answer

   Answer & explanation

   Correct: option A

   Gradient of $y = 3x - 1$ is $3$. Perpendicular gradient $= -\frac{1}{3}$ (negative reciprocal). Only option A has this gradient.

3. 3

   The quadratic $y = x^2 - 5x + 6$ has roots at:

   • A

     $x = 2$ and $x = 3$

   • B

     $x = -2$ and $x = -3$

   • C

     $x = 1$ and $x = 6$

   • D

     $x = 2$ and $x = -3$

   Check answer

   Answer & explanation

   Correct: option A

   Factorise: $(x-2)(x-3) = 0$, so $x = 2$ or $x = 3$. Check: $2 \times 3 = 6$ ✓, $2 + 3 = 5$ ✓.

4. 4

   How many real roots does $2x^2 + 3x + 5 = 0$ have?

   • A

     None

   • B

     One repeated root

   • C

     Two distinct roots

   • D

     Cannot tell without solving

   Check answer

   Answer & explanation

   Correct: option A

   Discriminant $= 3^2 - 4(2)(5) = 9 - 40 = -31 < 0$. Negative discriminant means no real roots.

5. 5

   The graph of $y = f(x)$ is translated to give $y = f(x - 2) + 3$. What is the translation vector?

   • A

     $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$

   • B

     $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$

   • C

     $\begin{pmatrix} 2 \\ -3 \end{pmatrix}$

   • D

     $\begin{pmatrix} -2 \\ -3 \end{pmatrix}$

   Check answer

   Answer & explanation

   Correct: option A

   $f(x - 2)$ shifts right by 2; $+3$ shifts up by 3. Translation vector $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$.

6. 6

   On a velocity-time graph, what does the area under the graph represent?

   • A

     Distance travelled

   • B

     Acceleration

   • C

     Speed

   • D

     Time

   Check answer

   Answer & explanation

   Correct: option A

   Area under a velocity-time graph = distance travelled. Gradient of a velocity-time graph = acceleration.

7. 7

   Which of the following is the equation of a circle with radius 7 centred at the origin?

   • A

     $x^2 + y^2 = 49$

   • B

     $x^2 + y^2 = 7$

   • C

     $x^2 + y^2 = 14$

   • D

     $(x+7)^2 + (y+7)^2 = 0$

   Check answer

   Answer & explanation

   Correct: option A

   $x^2 + y^2 = r^2 = 7^2 = 49$.

8. 8

   The graph of $y = \cos x$ is reflected in the $x$-axis. What is the equation of the new graph?

   • A

     $y = -\cos x$

   • B

     $y = \cos(-x)$

   • C

     $y = \cos x + 1$

   • D

     $y = -\cos(-x)$

   Check answer

   Answer & explanation

   Correct: option A

   Reflection in the $x$-axis gives $y = -f(x)$, so $y = -\cos x$.

Exam-style multi-part questions. Try each part on paper, then reveal the model answer to compare your working.

1. Q16 marks total

   Show all answers

   Scenario

   The straight line $L$ passes through the points $A(1, 5)$ and $B(4, -1)$.

   1. (a)

      Find the gradient of $L$.

      [2]
      Show answer

      Model answer · 2 marks

      $m = \dfrac{-1 - 5}{4 - 1} = \dfrac{-6}{3} = -2$

   2. (b)

      Find the equation of $L$, giving your answer in the form $y = mx + c$.

      [2]
      Show answer

      Model answer · 2 marks

      Using $A(1,5)$: $5 = -2(1) + c \implies c = 7$. So $y = -2x + 7$.

   3. (c)

      Find the equation of the line perpendicular to $L$ that passes through $A(1, 5)$.

      [2]
      Show answer

      Model answer · 2 marks

      Perpendicular gradient $= \frac{1}{2}$. Using $A$: $5 = \frac{1}{2}(1) + c \implies c = \frac{9}{2}$. So $y = \frac{1}{2}x + \frac{9}{2}$.

   Examiner note

   Many candidates correctly find the gradient in (a) but then make a sign error when finding c in (b). Always substitute a specific point to find c rather than guessing.

2. Q27 marks total

   Show all answers

   Scenario

   The diagram shows the graph of $y = x^2 - 4x + 3$.

   1. (a)

      Write down the coordinates of the $y$-intercept.

      [1]
      Show answer

      Model answer · 1 mark

      Set $x = 0$: $y = 3$. $y$-intercept is $(0, 3)$.

   2. (b)

      Find the roots of $y = x^2 - 4x + 3$.

      [2]
      Show answer

      Model answer · 2 marks

      Factorise: $(x-1)(x-3) = 0$, so $x = 1$ and $x = 3$. Roots at $(1, 0)$ and $(3, 0)$.

   3. (c)

      Find the coordinates of the turning point.

      [2]
      Show answer

      Model answer · 2 marks

      $x = -\dfrac{-4}{2(1)} = 2$. $y = 4 - 8 + 3 = -1$. Turning point is $(2, -1)$.

   4. (d)

      Using your answers, sketch the graph and label all key features.

      [2]
      Show answer

      Model answer · 2 marks

      U-shaped parabola through $(0,3)$, $(1,0)$, $(3,0)$ with minimum at $(2,-1)$ and line of symmetry $x = 2$.

3. Q35 marks total

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   Scenario

   A point $P(3, 4)$ lies on the circle $x^2 + y^2 = 25$.

   1. (a)

      Verify that $P$ lies on the circle.

      [1]
      Show answer

      Model answer · 1 mark

      $3^2 + 4^2 = 9 + 16 = 25$ ✓

   2. (b)

      Find the gradient of the radius $OP$ where $O$ is the origin.

      [1]
      Show answer

      Model answer · 1 mark

      Gradient of $OP = \dfrac{4 - 0}{3 - 0} = \dfrac{4}{3}$

   3. (c)

      Find the equation of the tangent to the circle at $P$. Give your answer in the form $ax + by = c$.

      [3]
      Show answer

      Model answer · 3 marks

      Tangent gradient $= -\dfrac{3}{4}$ (negative reciprocal). Using $P(3,4)$: $y - 4 = -\dfrac{3}{4}(x - 3)$. Multiply by 4: $4y - 16 = -3x + 9$. So $3x + 4y = 25$.

   Examiner note

   The shortcut ax + by = r² can be used directly for a circle centred at the origin: tangent at (3,4) on x²+y²=25 is 3x + 4y = 25.