Graphs
Algebra
- Crossover
- Non-calculator
- 8 questions
- 29 marks
Questions marked Higher are aimed at the Higher tier.
These are original practice questions, not a past paper. This site is independent and not affiliated with AQA or Pearson.
-
A straight line passes through $(0, 2)$ and $(3, 8)$. Find its gradient.
2 marks
-
A line has gradient $3$ and crosses the $y$-axis at $-1$. Write its equation.
2 marks
-
Complete the values of $y = x^{2} - 2x$ for $x = 0, 1, 2, 3$.
3 marks
-
The line in the diagram has equation $y = 2x + 1$. Write the coordinates of the $y$-intercept, and the value of $y$ when $x = 3$.
The line y = 2x + 1 3 marks
-
For the curve $y = x^{2} - 4$, find the coordinates where it crosses the axes.
4 marks
-
Does $(2, 7)$ lie on $y = 3x + 1$? Does $(1, 5)$ lie on $y = x^{2} + 3x + 1$?
4 marks
-
Water is poured at a steady rate. The depth-time graph is the straight line from $(0, 0)$ to $(10, 8)$, with time in minutes and depth in centimetres. Find the depth after $5$ minutes, the rate of increase, and the time taken to reach a depth of $6$ cm.
Depth of water against time 5 marks
-
For $y = x^{2} - 4x + 3$, find $y$ when $x = 0$, $x = 2$ and $x = 4$. Solve $y = 0$, and write down the turning point. Higher
6 marks
Mark scheme
-
Answer. $2$
- $\dfrac{8 - 2}{3 - 0}$
- $2$
-
Answer. $y = 3x - 1$
- The intercept is $-1$.
- $y = 3x - 1$
-
Answer. $0, -1, 0, 3$
- When $x = 1$, $y = 1 - 2 = -1$.
- The four values are $0, -1, 0, 3$.
-
Answer. $(0, 1)$, and $y = 7$
- The line crosses the $y$-axis at $(0, 1)$.
- $y = 2 \times 3 + 1 = 7$.
-
Answer. $(0, -4)$, $(2, 0)$ and $(-2, 0)$
- When $x = 0$, $y = -4$.
- $x^{2} - 4 = 0$ gives $x = \pm 2$.
-
Answer. Yes and yes
- $3 \times 2 + 1 = 7$, so the first point lies on the line.
- $1 + 3 + 1 = 5$, so the second point lies on the curve.
-
Answer. $4$ cm, $0.8$ cm/min, and $7.5$ minutes
- The midpoint of the line gives a depth of $4$ cm.
- Gradient $= \dfrac{8}{10} = 0.8$ cm/min.
- Time for $6$ cm $= \dfrac{6}{0.8} = 7.5$ minutes.
-
Answer. $y = 3, -1, 3$. Roots $x = 1$ and $x = 3$. Turning point $(2, -1)$.
- The three $y$ values are $3$, $-1$ and $3$.
- $(x - 1)(x - 3) = 0$, so $x = 1$ or $x = 3$.
- The turning point is halfway between the roots, at $(2, -1)$.