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Coordinate Geometry - Graphs of Functions (Quadratic, Cubic, Reciprocal)

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Quadratic graphs of the form y=ax2+bx+cy = ax^2 + bx + c are called parabolas. If a>0a > 0, the parabola opens upwards (U-shaped) and has a minimum point. If a<0a < 0, it opens downwards (n-shaped) and has a maximum point. The yy-intercept is always at (0,c)(0, c).

Graph of a quadratic function showing a U-shaped parabola.
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Cubic graphs of the form y=ax3+bx2+cx+dy = ax^3 + bx^2 + cx + d typically have an 'S' shape. They can have up to two turning points and always cross the xx-axis at least once.

Graph of a cubic function showing an S-shape with two turning points.
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Reciprocal graphs y=kxy = \frac{k}{x} create hyperbolas. They have two separate branches and asymptotic behavior, meaning the curve approaches the xx and yy axes but never touches them (for k/xk/x).

Graph of y = 1/x showing two hyperbolic branches in the first and third quadrants.
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Square reciprocal graphs y=kx2y = \frac{k}{x^2} are always positive (if k>0k > 0) and are symmetric about the yy-axis. They are often referred to as 'chimney' or 'volcano' graphs.

📐Formulae

General Quadratic: y=ax2+bx+cy = ax^2 + bx + c

Quadratic Formula: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Axis of Symmetry for Quadratic: x=−b2ax = -\frac{b}{2a}

General Cubic: y=ax3+bx2+cx+dy = ax^3 + bx^2 + cx + d

Reciprocal: y=kxy = \frac{k}{x} or y=kx2y = \frac{k}{x^2}

💡Examples

Problem 1:

Given the function f(x)=x2−4x+3f(x) = x^2 - 4x + 3, find the coordinates of the turning point (vertex).

Solution:

  1. Find the x-coordinate of the symmetry axis: x=−b/(2a)=−(−4)/(2×1)=2x = -b / (2a) = -(-4) / (2 \times 1) = 2.
  2. Substitute x=2x = 2 into the function: y=(2)2−4(2)+3=4−8+3=−1y = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1.
  3. Turning point is (2,−1)(2, -1).

Explanation:

The turning point of a quadratic is the minimum or maximum point. Using x=−b/2ax = -b/2a is the fastest way to find its location.

Problem 2:

Identify the horizontal and vertical asymptotes for the function y=3x−2+5y = \frac{3}{x-2} + 5.

Solution:

  1. Vertical Asymptote: Set denominator to zero: x−2=0⇒x=2x - 2 = 0 \Rightarrow x = 2.
  2. Horizontal Asymptote: As xx becomes very large, 3x−2\frac{3}{x-2} approaches 0, so yy approaches 5. y=5y = 5.

Explanation:

Asymptotes are lines that the curve approaches. A vertical asymptote occurs where the function is undefined (division by zero).

Problem 3:

Use the graph of y=x3−3xy = x^3 - 3x to estimate the solutions to x3−3x=1x^3 - 3x = 1.

Solution:

  1. Plot the curve y=x3−3xy = x^3 - 3x.
  2. Draw the horizontal line y=1y = 1 on the same grid.
  3. Identify the x-coordinates where the line and the curve intersect.

Explanation:

In IGCSE exams, 'solving graphically' means finding the intersection points between the function curve and a specific constant line or another function.

Problem 4:

Sketch the graph of y=4−x2y = 4 - x^2 for −3≤x≤3-3 \le x \le 3. State the coordinates of the maximum point and the xx-intercepts.

Graph of y = 4 - x^2 showing intercepts at -2 and 2 and a peak at 4.

Solution:

y=4−x2y = 4 - x^2 When x=0x = 0, y=4y = 4. So the yy-intercept (and maximum point) is (0,4)(0, 4). For xx-intercepts, set y=0y = 0: 4−x2=04 - x^2 = 0 x2=4x^2 = 4 x=±2x = \pm 2 Intercepts are (2,0)(2, 0) and (−2,0)(-2, 0).

Explanation:

This is a quadratic function where a=−1a = -1, so it is an inverted U-shape. The maximum point occurs at the vertex. We find intercepts by setting x=0x=0 and y=0y=0 respectively.

Problem 5:

Draw the graph of y=2xy = \frac{2}{x} for xx values from 0.50.5 to 44. Use the graph to solve 2x=1.5\frac{2}{x} = 1.5.

Graph of y = 2/x with a horizontal line at y = 1.5 intersecting the curve.

Solution:

Points for the graph: x=0.5,y=4x = 0.5, y = 4 x=1,y=2x = 1, y = 2 x=2,y=1x = 2, y = 1 x=4,y=0.5x = 4, y = 0.5 Drawing a horizontal line at y=1.5y = 1.5 gives: 2x=1.5\frac{2}{x} = 1.5 x=21.5=1.33x = \frac{2}{1.5} = 1.33

Explanation:

Create a table of values to plot the curve. To solve the equation graphically, identify the xx-coordinate where the curve intersects the line y=1.5y = 1.5.