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

Grade 10IGCSE

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

🔑Concepts

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Quadratic Functions (y=ax2+bx+cy = ax^2 + bx + c): Produce a U-shaped (parabola) if a>0a > 0 or an n-shaped curve if a<0a < 0. They are symmetrical about the vertex.

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Cubic Functions (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 at least one x-intercept.

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Reciprocal Functions (y=k/xy = k/x): Known as hyperbolas. They have two branches and asymptotes (lines the graph approaches but never touches), usually the x and y axes.

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Asymptotes: For y=k/xy = k/x, the vertical asymptote is x=0x = 0 and the horizontal asymptote is y=0y = 0.

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Roots and Intercepts: The x-intercepts (roots) are found where y=0y = 0. The y-intercept is found where x=0x = 0.

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Graphical Solutions: To solve f(x)=kf(x) = k, find the x-coordinates where the horizontal line y=ky = k intersects the curve y=f(x)y = f(x).

📐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.