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Find constants from gradient facts

Find unknown constants from gradient and point facts

  • In 4 of 20 recent exams
  • Paper 1 or 2
  • 2 to 6 marks
How it is set in the exam

Paper 1 (no calculator) (Paper 2 (calculator) with a GDC system), 2–6 marks. Typical.

Spot it in the exam
Exam phrases to spot (4)
  • “the tangent at x=2x = 2 is y=7x−10y = 7x - 10; find aa and bb”

  • “the gradient of the curve at (2,5)(2, 5) is 33”

  • “ff has a local minimum at (2,−1)(2, -1); find aa and bb”

  • “find the value of pp and the value of qq”

In disguise (3)
  • "the curves touch at x=1x = 1": equal values and equal gradients there
  • "a turning point on the y-axis": f′(0)=0f'(0) = 0
  • "the normal at x=2x = 2 is parallel to …": convert to the tangent's gradient first
Looks like it, but isn't (3)
  • f′(x)f'(x) is given, with a point on the graph: integrate (Find f from f′ and a point
  • the constants are known and the point is wanted: Find the point of contact
  • a line with an unknown is a tangent to a parabola, and no point is given: Two graphs
Worked example

One tangent line, two facts

Paper 1 (no calculator)
Question

The line y=7x−10y = 7x - 10 is a tangent to the curve y=ax2+bxy = ax^2 + \dfrac{b}{x} at the point where x=2x = 2. Find the value of aa and the value of bb. [6]

Try it yourself first, then compare.

Now try one like it

Write your working on paper or on screen. Paper Zero marks it step by step, like the worked example, and shows where each mark went. Free, with no account.

A handwritten page as Paper Zero marks it: ticks at the two lines that earned a mark, a cross on line 3 where +7 was copied as −7, and a short note beside that line.123456f′(x) = 4x − 124x − 12 = 0, so x = 3f(3) = 2(3)² − 12(3) − 7= 18 − 36 − 7 = −25minimum at (3, −25)M1 ✓A1 ✓A1 ✗11The +7 became −7: 18 − 36 + 7 = −11.
Steps
  1. Count

    the unknowns and the facts. Split every compound fact into its equations (table below): "the tangent at x=2x = 2 is y=7x−10y = 7x - 10" is two facts.

  2. Write one equation per fact.

    Differentiate first if a fact is about a gradient. When a tangent line is given, take the point's yy from the line.

  3. Solve

    , starting with the equation that has the fewest unknowns. On , write every equation down, then use the GDC (solve simultaneous linear equations).

  4. Check

    by substituting back into every fact.

    FactEquation(s)
    (x1,y1)(x_1, y_1) is on the curvef(x1)=y1f(x_1) = y_1
    the gradient at x=x1x = x_1 is mmf′(x1)=mf'(x_1) = m
    the tangent at x=x1x = x_1 is y=mx+cy = mx + cf′(x1)=mf'(x_1) = m and f(x1)=mx1+cf(x_1) = mx_1 + c (two facts)
    the normal at x=x1x = x_1 has gradient nnf′(x1)=−1nf'(x_1) = -\frac{1}{n}
    two curves touch (share a tangent) at x=x1x = x_1f(x1)=g(x1)f(x_1) = g(x_1) and f′(x1)=g′(x1)f'(x_1) = g'(x_1)
    a stationary point (turning point, maximum, minimum) at (x1,y1)(x_1, y_1)f′(x1)=0f'(x_1) = 0 and f(x1)=y1f(x_1) = y_1
    a point of inflexion at x=x1x = x_1f′′(x1)=0f''(x_1) = 0
Common mistakes
  • Using a tangent line as one fact.

    "The tangent at x=2x = 2 is y=7x−10y = 7x - 10" gives two equations: the point and the gradient.

  • Taking the yy-value from the curve

    before the constants are known. When the tangent line is given, the point's yy comes from the line.

  • Using only f′(x1)=0f'(x_1) = 0 for a stationary point (x1,y1)(x_1, y_1).

    f(x1)=y1f(x_1) = y_1 is the second equation.

More mistakes (3)
  • Differentiating a constant term bb as if it were bxbx.

    It differentiates to 00.

  • Not converting a normal's gradient.

    A normal gradient nn means f′(x1)=−1nf'(x_1) = -\frac{1}{n}.

  • Rounding aa and bb

    before using them in a later part ().

Next typeThe tangent or normal meets something