Grade 12 Maths · Calculus

Find equation of a tangent

Find the equation of a tangent to a curve using the derivative as its gradient at the point where it touches (for example m = f′(2)).

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Equation of Tangent Calculus Grade 12Kevinmathscience

How to Find the Equation of a Tangent | Calculus Grade 12 | NSC Paper 1Sancho Maths

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Question 8.3 · May/June 2024 · Paper 1 · National (DBE) · 4 marks

8.3 The equation of the tangent to

f (x) = 3 x^{2} + b x + c

x = 1

is given by

y = 9 x - 9

. Determine the values of

b

and

c

.(4)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: find equation of a tangent

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3. Question 6.6 · May/June 2024 · Paper 1 · National (DBE) · 5 marks
The graphs of $f (x) = x^{2} - 2 x - 3$ and $g (x) = m x + c$ are drawn below. D and E are the $x$ -intercepts and P is the $y$ -intercept of $f$ . The turning point of $f$ is T $(1; - 4)$ . The graphs of $f$ and $g$ intersect at P and E.
6.6 Given: $k (x) = g (x) - n$ . Determine $n$ if $k$ is a tangent to $f$ .(5)
Show the memo (5 ticks)Memo
1. ✓
  Equate derivative of f to gradient of g $2 x - 2 = 1$
  1 mark · Find equation of a tangent
2. ✓
  Find x-coordinate of tangency $x = 1, 5$
  1 mark · Find equation of a tangent
3. ✓
  Find y-coordinate on f $y = (1, 5)^{2} - 2 (1, 5) - 3 = - 3, 75$
  1 mark · Find equation of a tangent
4. ✓
  Substitute point into k(x) equation $- 3, 75 = 1, 5 - 3 - n$
  1 mark · Find equation of a tangent
5. ✓
  Solve for n $n = 2, 25$
  1 mark · Find equation of a tangent
4. Question 6.4 · November 2024 · Paper 1 · National (DBE) · 5 marks
In the diagram below, the graphs of $f (x) = - x^{2} + 4 x + 5$ and $g$ , a straight line, are drawn. $C (3; 8)$ is a point of intersection of $f$ and $g$ . $E H$ is drawn parallel to the y-axis, with $E$ a point on $f$ and $H$ a point on $g$ .
6.4 Given: $k (x) = f (x + m) = - x^{2} - 2 m x - m^{2} + 4 x + 4 m + 5$ . Determine the value of $m$ such that $g$ is a tangent to $k$ .(5)
Show the memo (3 ticks)Memo
1. ✓
  METHOD 1: Using Derivative $k^{'} (x) = - 2 x - 2 m + 4$
  1 mark · Differentiate using the power rule
2. ✓
  Equate derivative to gradient of g $k^{'} (x) = 2$
  $- 2 x - 2 m + 4 = 2$
  $- 2 x = 2 m - 2$
  $x = 1 - m$
  1 mark · Find equation of a tangent
3. ✓
  Substitute and solve $k (1 - m) = g (1 - m)$
  $- (1 - m)^{2} - 2 m (1 - m) - m^{2} + 4 (1 - m) + 4 m + 5 = 2 (1 - m) + 2$
  Simplify: $- 2 m - 4 = 0$
  $m = - 2$
  3 marks · Find equation of a tangent
5. Question 8.3 · May/June 2025 · Paper 1 · National (DBE) · 6 marks
8.3 Given that $y = 4 x - 14$ is a common tangent to $f (x) = 2 x^{2} - 4 x - 6$ and $g (x) = a x^{2} + b x - 18$ , calculate the values of $a$ and $b$ .(6)
Show the memo (6 ticks)Memo
1. ✓
  Equate the derivative of f(x) to the gradient of the tangentThe gradient of the tangent $y = 4 x - 14$ is $m = 4$ .
  $f^{'} (x) = 4 x - 4$
  $∴ 4 x - 4 = 4$
  1 mark · Find equation of a tangent
2. ✓
  Solve for the x-coordinate of the point of tangency $4 x = 8$
  $x = 2$
  1 mark · Find equation of a tangent
3. ✓
  Calculate the y-coordinate at the point of tangencySubstitute $x = 2$ into the tangent equation:
  $y = 4 (2) - 14 = - 6$
  Point of tangency is $(2; - 6)$
  1 mark · Find equation of a tangent
4. ✓
  Use the point of tangency to form an equation with g(x)Since it is a common tangent, $g (x)$ also passes through $(2; - 6)$ .
  $g (2) = a (2)^{2} + b (2) - 18 = - 6$
  $4 a + 2 b - 18 = - 6$
  $4 a + 2 b = 12$ ... (Equation 1)
  1 mark · Find the equation of a parabola
5. ✓
  Use the gradient to form a second equation with g'(x)The gradient of $g (x)$ at $x = 2$ must also be $4$ .
  $g^{'} (x) = 2 a x + b$
  $g^{'} (2) = 2 a (2) + b = 4$
  $4 a + b = 4$ ... (Equation 2)
  1 mark · Find equation of a tangent
6. ✓
  Solve the equations simultaneously for a and bSubtract (2) from (1):
  $(4 a + 2 b) - (4 a + b) = 12 - 4$
  $b = 8$
  Substitute $b = 8$ into (2):
  $4 a + 8 = 4$
  $4 a = - 4$
  $a = - 1$
  1 mark · Solve simultaneous equations by elimination
6. Question 9.2 · September 2024 · Paper 1 · Provincial · 2 marks
9.2 Given: $h (x) = 4 x^{3} + 5 x$ . Substantiate whether it is possible to draw a tangent to the graph of h that has a negative gradient.(2)
Show the memo (2 ticks)Memo
1. ✓
  Derivative $h^{'} (x) = 12 x^{2} + 5$
  1 mark · Differentiate using the power rule
2. ✓
  ExplanationAlways positive
  1 mark · Find equation of a tangent

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