Worksheets · Foundation and Higher

Quadratic graphs, roots and turning points

8 exam-style questions, grades 3 to 6. Worked solutions and the marks are on the last page.

Question 1Non-calculator · 2 marks
Here is the equation of a curve: $y = x^2 + 1$ .
(a) Work out the value of $y$ when $x = -2$ . (1)
(b) Which point is the turning point of the curve? (1)
1. $(0, 1)$
2. $(1, 0)$
3. $(0, 0)$
Question 2Non-calculator · 1 mark
(a) The graph of y = (x + 1)(x $-$ 5) crosses the x-axis at two points. What are the x-coordinates of these points? (1)
1. 1 and $-5$
2. $-1$ and 5
3. 2 and $-9$
4. 0 and $-5$
Question 3Non-calculator · 5 marks
The curve has equation $y=(x-3)(x-5)$ .
(a) Find the coordinates where the curve crosses the x-axis. (2)
(b) Find the coordinates of the turning point. (3)
Question 4Non-calculator · 5 marks
The curve $y = x^2 - 6x + 5$ is drawn for $0 \le x \le 6$ .
(a) Find the coordinates of the point where the curve crosses the y-axis. (1)
(b) Factorise $x^2 - 6x + 5$ and hence write down the roots of $x^2 - 6x + 5 = 0$ . (2)
(c) Write down the equation of the line of symmetry of the curve. (2)
Question 5Non-calculator · 4 marks
A ball is thrown from ground level. Its height, h metres, after t seconds is $h = 12t - 3t^2$ for $0 \le t \le 4$ .
(a) Find the two times when the ball is at ground level. (2)
(b) Find the greatest height of the ball. (2)
Question 6Non-calculator · 4 marks
A quadratic curve crosses the x-axis at (−1, 0) and (5, 0) and passes through (0, −5).
(a) Find the equation of the curve in the form $y = x^2 + bx + c$ . (2)
(b) Find the coordinates of the turning point. (2)
Question 7Non-calculator · 4 marks
The curve $y = x^2 + 2x + 4$ has its turning point at (−1, 3).
(a) Explain why the equation $x^2 + 2x + 4 = 0$ has no real solutions. (1)
(b) Write down the value of k for which the line y = k touches the curve at exactly one point. (1)
(c) Find the value of y on the curve when x = 2. (2)
Question 8Non-calculator · 5 marks
The curve $y = (x - 3)(x + 5)$ is a parabola.
(a) Write down the coordinates of the points where the curve crosses the $x$ -axis. (1)
(b) Find the coordinates of the turning point of the curve. (3)
(c) For which value of $k$ does the equation $(x - 3)(x + 5) = k$ have exactly one solution? (1)

Worked solutions and marks

Question 1

(a) $5$

$(-2)^2 + 1 = 4 + 1 = 5$ .

B1 The correct answer, $5$ .

(b) $(0, 1)$

$x^2$ is smallest when $x = 0$ , so the lowest point is $(0, 1)$ .

B1 The correct answer, $(0, 1)$ .

Question 2

(a) $-1$ and 5

At an x-axis crossing y = 0.
Set each factor to zero: x + 1 = 0 or x $-$ 5 = 0, so x = $-1$ or x = 5.

B1 Correct answer: $-1$ and 5

Question 3

(a) $(3,0)$ and $(5,0)$

Set y equal to zero and then set each factor equal to zero.
$(x-3)(x-5)=0$
Therefore $(3,0)$ and $(5,0)$ .

M1 Set y equal to zero and then set each factor equal to zero.
A1 Correct answer: $(3,0)$ and $(5,0)$

(b) $(4,-1)$

The axis of symmetry is halfway between the roots.
$\frac{3+5}{2}$
Substitute the midpoint into the curve equation.
$(4-3)(4-5)$
Therefore $(4,-1)$ .

M1 The axis of symmetry is halfway between the roots.
M1 Substitute the midpoint into the curve equation.
A1 Correct answer: $(4,-1)$

Question 4

(a) $(0,5)$

Therefore $(0,5)$ .

B1 Correct answer: $(0,5)$

(b) $1, 5$

Factorise the quadratic.
$(x-1)(x-5)$
Therefore $1, 5$ .

M1 Factorise the quadratic.
A1 Correct answer: $1, 5$

(c) $x=3$

The line of symmetry is halfway between the roots.
$\frac{1+5}{2}$
Therefore $x=3$ .

M1 The line of symmetry is halfway between the roots.
A1 Correct answer: $x=3$

Question 5

(a) $0, 4$

Set h = 0 and factorise.
$3t(4-t)=0$
Therefore $0, 4$ .

P1 Set h = 0 and factorise.
A1 Correct answer: $0, 4$

(b) $12$ m

The greatest height is halfway between the two times, at t = 2.
$12\times 2-3\times 2^{2}$
Therefore $12$ m.

P1 The greatest height is halfway between the two times, at t = 2.
A1 Correct answer: $12$ m

Question 6

(a) $y=x^2-4x-5$

Write the curve using its roots.
$(x+1)(x-5)$
Therefore $y=x^2-4x-5$ .

M1 Write the curve using its roots.
A1 Correct answer: $y=x^2-4x-5$

(b) $(2,-9)$

The turning point is at x = 2, halfway between the roots.
$2^{2}-4\times 2-5$
Therefore $(2,-9)$ .

M1 The turning point is at x = 2, halfway between the roots.
A1 Correct answer: $(2,-9)$

Question 7

(a) The lowest point of the curve has y = 3, which is above 0, so the curve never meets the x-axis.

The lowest point of the curve has y = 3, which is above 0, so the curve never meets the x-axis.

C1 Correct conclusion with supporting reasoning: The lowest point of the curve has y = 3, which is above 0, so the curve never meets the x-axis.

(b) $3$

Therefore $3$ .

B1 Correct answer: $3$

(c) $12$

Substitute x = 2.
$2^{2}+2\times 2+4$
Therefore $12$ .

M1 Substitute x = 2.
A1 Correct answer: $12$

Question 8

(a) $(3, 0)$ and $(-5, 0)$

$y = 0$ when $x - 3 = 0$ or $x + 5 = 0$ : $x = 3$ or $x = -5$ .

B1 $x = 3$ and $x = -5$ .

(b) $(-1, -16)$

The turning point is on the line of symmetry, halfway between the roots.
$x = \frac{3 + (-5)}{2} = -1$
Substitute.
$y = (-1 - 3)(-1 + 5) = (-4)(4) = -16$

P1 Finding the $x$ -coordinate, $-1$ , halfway between the roots.
P1 Substituting $x = -1$ into the equation.
A1 The correct answer, $(-1, -16)$ .

(c) $k = -16$

The line $y = k$ touches the curve at exactly one point only at the turning point, where $y = -16$ .

B1 The correct answer, $k = -16$ .

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Quadratic graphs, roots and turning points

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