Printable · GCSE Higher · ages 14-16

Roots, intercepts and turning points of quadratics worksheet — GCSE Higher

Fifteen questions on "roots, intercepts and turning points of quadratics" — DfE statement A11. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.

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Roots, intercepts and turning points of quadratics worksheet — GCSE Higher

Maths · GCSE Higher · A11 · Roots, intercepts and turning points of quadratics · 15 questions

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1.By completing the square, show that the curve y = x² − 14x + 50 never crosses the x-axis. Which statement gives the correct reason?
y = x² − 14x + 50
(a)The turning point is above the x-axis at (7, −1)
(b)50 is positive, so y is always positive
(c)The discriminant is positive, so y is never zero
(d)(x − 7)² + 1 is at least 1 for every x, so y is never 0
2.A curve has equation y = x² − 9. Which statement about its graph is correct?
y = x² − 9
(a)It crosses the x-axis only once, at x = 9.
(b)It does not cross the x-axis, since −9 is negative.
(c)It crosses the x-axis at x = 9 and x = −9.
(d)It crosses the x-axis at x = 3 and x = −3.
3.The graph of y = x² − 7x + 2 crosses the x-axis at two points. One root, read from the graph, is approximately x = 0.30. Using the fact that the sum of the two roots of x² − 7x + 2 = 0 is 7, estimate the other root, correct to 2 decimal places.
y = x² − 7x + 2
(a)x ≈ 6.70
(b)x ≈ 6.30
(c)x ≈ 7.30
(d)x ≈ 0.70
4.By completing the square, find the turning point of the curve y = x² − 10x + 30.
y = x² − 10x + 30
(a)x = 10, y = −70
(b)x = 5, y = −5
(c)x = 5, y = 5
(d)x = −5, y = 5
5.The curve y = 2x² − 12x + 7 has a minimum point. By completing the square, find the value of x at which the minimum occurs.
y = 2x² − 12x + 7
(a)6
(b)−3
(c)−12
(d)3
6.By completing the square, find the turning point of the curve y = 2x² − 8x + 3.
y = 2x² − 8x + 3
(a)x = 4, y = −29
(b)x = 2, y = −5
(c)x = −2, y = −5
(d)x = 2, y = 5
7.By completing the square, find the turning point of the curve y = 3x² + 12x + 7.
y = 3x² + 12x + 7
(a)x = −2, y = 5
(b)x = −4, y = −41
(c)x = 2, y = −5
(d)x = −2, y = −5
8.Write y = x² + 12x + 40 in the form (x + a)² + b, and hence write down the minimum value of y.
y = x² + 12x + 40
(a)−6
(b)36
(c)4
(d)40
9.A graph has equation y = x² − 6x + 5. A student says its turning point has x-coordinate 6, because that's the coefficient of x. Which statement corrects the student's mistake?
y = x² − 6x + 5
(a)The student is correct, because the turning point's x-coordinate always equals the coefficient of x.
(b)The turning point has x-coordinate −6, because the sign of the coefficient must be reversed.
(c)The turning point has x-coordinate 5, because that is the larger of the equation's two roots.
(d)The roots of x² − 6x + 5 = 0 are x = 1 and x = 5 (since it factorises to (x − 1)(x − 5)), so by symmetry the turning point has x-coordinate 3, not 6.
10.Which of these quadratic graphs does NOT cross the x-axis at all?
(a)y = (x − 2)² + 3
(b)y = (x + 4)(x − 1)
(c)y = (x − 2)(x + 3)
(d)y = x² − 9
11.The height, h metres, of a ball t seconds after a stopwatch is started follows h = (t − 1)(5 − t) for 1 ≤ t ≤ 5, where h = 0 means the ball is at ground level. Work out the two times at which the ball is at ground level.
(a)t = 1 second and t = 4 seconds
(b)t = 1 second and t = −5 seconds
(c)t = 1 second and t = 5 seconds
(d)t = −1 second and t = 5 seconds
12.The graph of y = x² − 5x + 6 crosses the x-axis at two points. By factorising, work out the x-coordinates of these two points.
y = x² − 5x + 6
(a)x = −1 and x = −6
(b)x = 2 and x = −3
(c)x = 2 and x = 3
(d)x = −2 and x = −3
13.The curve y = −x² + 6x − 5 has a maximum point. Use completing the square to find its coordinates.
y = −x² + 6x − 5
(a)x = 3, y = −4
(b)x = 3, y = 4
(c)x = 6, y = 31
(d)x = −3, y = 4
14.A quadratic graph has roots at x = −3 and x = 5, and it crosses the y-axis at (0, −15). Work out the equation of the curve in the form y = (x − a)(x − b).
(a)y = (x − 3)(x − 5)
(b)y = (x − 3)(x + 5)
(c)y = (x + 3)(x + 5)
(d)y = (x + 3)(x − 5)
15.A ball's height, h metres, t seconds after being thrown follows h = (t − 1)(9 − t). Given that the ball is at ground level at t = 1 and t = 9, work out at what time t the ball reaches its maximum height, using symmetry.
(a)t = 4
(b)t = 8
(c)t = 10
(d)t = 5

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