Most tested Papers 1 & 2

Graphs of Functions: Sketching, Transformations and Intersections (TMUA)

Recognising a function's shape from its equation turns algebra into geometry: real roots become x-axis crossings, a pair of simultaneous equations becomes a set of intersection points, and a transformation slides or stretches a curve you already know into a new one. Under no-calculator, multiple-choice conditions this fluency lets you reject wrong options by shape, symmetry or intercept alone, often before doing any algebra.

Part of the TMUA syllabus — revision for the Test of Mathematics for University Admission (TMUA), the UAT-UK maths test used by Cambridge, Oxford, Imperial, UCL, LSE, Warwick and Durham. No calculator; multiple choice.

Key points

  • Standard shapes to know on sight: linear y = mx + c (straight line); quadratic y = ax2 + bx + c (parabola, U-shaped if a > 0, n-shaped if a < 0); cubic (zero or two turning points, i.e. a local max and min together or neither); reciprocal y = 1/x (two branches with the axes as asymptotes); exponential y = a^x (through (0,1), rising if a > 1); square-root y = √(x); modulus y = |x| (V-shape); and the periodic sin, cos, tan curves.
    upper half of a sideways parabola, x ≥ 0
  • Outside-the-bracket transformations behave as expected: y = f(x) + a shifts UP by a, and y = a × f(x) is a vertical stretch of factor a (a reflection in the x-axis when a is negative).
  • Inside-the-bracket transformations act in the OPPOSITE sense to intuition: y = f(x + a) shifts LEFT by a, and y = f(ax) is a horizontal stretch of factor 1/a (a squash when a > 1).
  • For a line y = mx + c, m is the gradient (steepness and sign of the slope) and c is the y-intercept; parallel lines share the same m, and a negative m falls left-to-right.
  • Completed-square form y = a(x + b)2 + c reveals the parabola's vertex at (-b, c) with no calculus, while a fixes its width and whether it opens up or down.
  • Differentiation reads off shape: dy/dx > 0 means increasing, dy/dx < 0 means decreasing, and dy/dx = 0 marks a stationary point (classify by the sign change of the gradient either side).
  • Real roots are exactly the x-axis crossings; for a quadratic the discriminant b2 - 4ac decides the count: positive gives two, zero gives one repeated root (a tangent touch), negative gives none.
  • The intersections of two graphs are the solutions of their simultaneous equations: substitute to reach a single equation, and the number of real solutions equals the number of crossing points (tangency corresponds to a repeated solution).

Diagram

xy-6-303-404y = x^2y = (x+3)^2 - 4(-3, -4)
A translation: y = x2 (dashed) becomes y = (x + 3)2 - 4 (solid) - shifted 3 left and 4 down.

Formulae

y = a(x + b)2 + c has vertex (-b, c)

Reading a parabola's turning point and line of symmetry straight from completed-square form, no differentiation needed.

discriminant = b2 - 4ac

Counting the real roots of ax2 + bx + c = 0, or forcing tangency between a line and a curve by setting it to 0.

y = f(x + a) shifts left a; y = f(x) + a shifts up a

Applying single translations and remembering that the inside change is 'opposite', the outside change is 'as expected'.

y = f(ax) stretches horizontally 1/a; y = a × f(x) stretches vertically a

Applying stretches and squashes, and spotting reflections when the factor is negative (y = -f(x) flips in the x-axis, y = f(-x) flips in the y-axis).

m = (y2 - y1)/(x2 - x1)

Finding the gradient of a line through two known points before writing it as y = mx + c.

Definitions

Stationary point
A point on a curve where the gradient is zero, i.e. dy/dx = 0. It may be a local maximum, a local minimum or a point of inflection, decided by how the gradient's sign changes around it.
Asymptote
A straight line that a curve approaches arbitrarily closely but never actually reaches, such as the x- and y-axes for y = 1/x or the horizontal axis for a decaying exponential.
Discriminant
The quantity b2 - 4ac formed from a quadratic ax2 + bx + c. Its sign alone determines whether the parabola crosses the x-axis twice, touches it once, or misses it entirely.
Vertex (turning point)
The single lowest or highest point of a parabola. In completed-square form y = a(x + b)2 + c it sits exactly at (-b, c), giving the curve's line of symmetry x = -b.

Worked examples

1

The curve y = x2 is transformed into y = (x + 3)2 - 4. Give the coordinates of the new vertex and the x-values where the new curve meets the x-axis.

  1. 1

    Compare with a(x + b)2 + c:

    a = 1, b = 3, c = -4
  2. 2

    Read off the turning point:

    vertex = (-3, -4)
  3. 3
    Put y = 0 and isolate the square:
    (x + 3)2 = 4
  4. 4

    Take the square root of both sides:

    x + 3 = ± 2
  5. 5

    Solve each case:

    x = -5 or x = -1

Answer: The vertex is (-3, -4), and the curve crosses the x-axis at x = -5 and x = -1.

2

Find the value of k for which the line y = 2x + k is a tangent to the curve y = x2 + 3x + 5 (that is, it meets the curve at exactly one point).

  1. 1

    Set the line equal to the curve:

    2x + k = x2 + 3x + 5
  2. 2

    Collect everything on one side:

    0 = x2 + x + 5 - k
  3. 3

    Tangency means one repeated root:

    discriminant = 0.

  4. 4

    Apply b2 - 4ac with a = 1, b = 1, c = 5 - k:

    1 - 4(5 - k) = 0.

  5. 5

    Simplify the constant:

    4k - 19 = 0
  6. 6

    Solve for k:

    k = 19/4

Answer: k = 19/4 (equivalently 4.75).

Common mistakes

  • ×Reversing the translation direction: y = f(x + a) moves the graph LEFT by a, not right, because the inside-bracket sign is opposite to the shift.
  • ×Confusing stretch factors: y = f(2x) squashes the curve horizontally to half its width (factor 1/2); it does not stretch it by 2.
  • ×Assuming every quadratic crosses the x-axis: if b2 - 4ac < 0 there are no real roots and the parabola misses the axis entirely, even though the curve still exists.
  • ×Misreading the vertex sign from a(x + b)2 + c: the x-coordinate of the turning point is -b, not b.
  • ×Applying combined transformations in the wrong order: an inside shift followed by a stretch generally lands on a different curve than the stretch followed by the shift.

No-calculator tips

  • ✓Eliminate answer options by shape first: check the direction of opening (sign of a), the intercepts and any asymptotes before committing to algebra.
  • ✓When a question only asks HOW MANY roots or intersections, use just the sign of the discriminant; you rarely need to finish the arithmetic.
  • ✓Substitute one easy point, usually x = 0, to test a transformed equation and knock out mismatched options instantly.
  • ✓Take parabola turning points directly from completed-square form rather than differentiating and solving dy/dx = 0.
  • ✓For a tangent, set line equal to curve and demand discriminant = 0; this is faster and cleaner than solving the quadratic in full.

Test yourself

Original practice questions, no calculator. Work each out before revealing the answer.

Q1.The quadratic f(x) = 3x2 - 12x + 5 is written in the completed-square form a(x + b)2 + c. What are the coordinates of the vertex (turning point) of y = f(x)?

  • A. (2, 1)
  • B. (-2, -7)
  • C. (2, -7)
  • D. (2, 5)
  • E. (-2, 1)
Show answer

Answer: C — (2, -7)

3(x2 - 4x) + 5 = 3(x - 2)2 - 12 + 5 = 3(x - 2)2 - 7, so the vertex is (2, -7). The trap (2, 1) comes from forgetting to multiply the -4 inside the bracket by the factor 3.

Q2.The line y = 2x + c is a tangent to the curve y = x2 + 3. What is the value of c?

  • A. c = 4
  • B. c = 2
  • C. c = 1
  • D. c = -2
Show answer

Answer: B — c = 2

Setting x2 + 3 = 2x + c gives x2 - 2x + (3 - c) = 0; tangency means the discriminant is 0, so 4 - 4(3 - c) = 0, giving c = 2. (c = 1 is the constant of the perfect square x2 - 2x + 1, not the answer.)

Q3.The graph of y = f(x) is transformed into the graph of y = f(x + 2) - 3. Which single description is correct?

  • A. a translation 2 to the left and 3 down
  • B. a translation 2 to the right and 3 down
  • C. a translation 2 to the left and 3 up
  • D. a translation 2 to the right and 3 up
  • E. a translation 3 to the left and 2 down
Show answer

Answer: A — a translation 2 to the left and 3 down

Replacing x by x + 2 shifts the graph 2 to the LEFT (changes inside the function act in the opposite direction), and the outside -3 shifts it 3 down. The common error is reading f(x + 2) as a shift to the right.

Read this topic in the official UAT-UK TMUA content specification →

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