Grade 10 Maths · Functions

Find where two graphs intersect

Find where two graphs meet by setting their equations equal and solving (for example x² = x + 2 gives the points (2; 4) and (−1; 1)).

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Parabola + Straight Line| Grade 11 FunctionsKevinmathscience

How to find the points of intersection of a straight line and a parabola.Maths Unmasked

Functions Parabola Grade11: L9 Point of Intersection of 2 graphsOnline Maths by Miss Pythagoras

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Question 4.5 · May/June 2025 · Paper 1 · National (DBE) · 3 marks

The graph of

f (x) = \frac{4}{x - 3} + 4

is drawn below. M is the point where the asymptotes of

f

intersect. C and D are the x- and y-intercepts respectively of

f

. A is the point on

f

that is closest to M. Diagram for this question

4.5 Calculate the coordinates of A.(3)

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

More past paper questions: find where two graphs intersect

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3. Question 1.7 · June 2024 · Paper 2 · Provincial · 4 marks
In the diagram below, $A (- 2; 2)$ , $B (- 4; - 6)$ , $C (4; - 2)$ and $E (6; y)$ are the vertices of a quadrilateral with $A E ∣∣ BC$ . $D$ lies on $BC$ such that $A D ⊥ BC$ and $A C$ is drawn. $M (x; 2)$ is a point on line $EC$ .
1.7 Show that the coordinates of $D$ are $(0, 8; - 3, 6)$ .(4)
Show the memo (4 ticks)Memo
1. ✓
  Determine equation of BC $y - (- 2) = \frac{1}{2} (x - 4) ⟹ y = \frac{1}{2} x - 4$
  1 mark · Find the equation of a line
2. ✓
  Set equations equal (Intersection of AD and BC) $- 2 x - 2 = \frac{1}{2} x - 4$
  1 mark · Find where two graphs intersect
3. ✓
  Solve for xMultiply by 2: $- 4 x - 4 = x - 8$
  $- 5 x = - 4 ⟹ x = 0, 8$ (or $\frac{4}{5}$ )
  1 mark · Find where two graphs intersect
4. ✓
  Solve for y $y = - 2 (0, 8) - 2 = - 1, 6 - 2 = - 3, 6$
  $D (0, 8; - 3, 6)$
  1 mark · Find where two graphs intersect
4. 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
5. 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
6. Question 4.1.5 · March 2018 · Provincial · 4 marks
4.1 $A (p; q), B (2; 4), C (3; 0)$ and $D (- 1; - 2)$ are the vertices of parallelogram ABCD. O is the origin and OE is perpendicular to AB.
4.1.5 Calculate the coordinates of E.(4)
Show the memo (4 ticks)Memo
1. ✓
  Equation of AB $y - 4 = \frac{1}{2} (x - 2)$
  $y = \frac{1}{2} x + 3$
  1 mark · Find the equation of a line
2. ✓
  Equate equations $- 2 x = \frac{1}{2} x + 3$
  $- \frac{5}{2} x = 3$
  1 mark · Find where two graphs intersect
3. ✓
  Solve for x $x = - \frac{6}{5}$ or $- 1, 2$
  1 mark · Find where two graphs intersect
4. ✓
  Solve for y $y = - 2 (- 1, 2) = 2, 4$ or $\frac{12}{5}$
  1 mark · Find where two graphs intersect

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