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Algebra - Sketching Curves

Grade 9IGCSE

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

🔑Concepts

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Linear Graphs: The general equation is y=mx+cy = mx + c. These are straight lines where mm represents the gradient (slope) and cc represents the yy-intercept.

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Quadratic Graphs: Equations in the form y=ax2+bx+cy = ax^2 + bx + c. These form a parabola. If a>0a > 0, the curve is a 'U' shape (minimum point). If a<0a < 0, the curve is an 'n' shape (maximum point).

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Intercepts: To sketch any curve, find the yy-intercept by setting x=0x = 0 and the xx-intercepts (roots) by setting y=0y = 0.

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Cubic Graphs: Equations in the form y=ax3+bx2+cx+dy = ax^3 + bx^2 + cx + d. They generally have an 'S' shape and can have up to two turning points and three xx-intercepts.

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Reciprocal Graphs: Equations like y=kxy = \frac{k}{x} where x≠0x \neq 0. These graphs have asymptotes (lines the curve approaches but never touches), typically the xx-axis (y=0y=0) and yy-axis (x=0x=0).

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Exponential Graphs: Equations in the form y=axy = a^x where a>0a > 0. These curves always pass through (0,1)(0, 1) if not transformed and grow rapidly for a>1a > 1.

📐Formulae

y=mx+cy = mx + c

y=ax2+bx+cy = ax^2 + bx + c

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

x=−b2ax = -\frac{b}{2a}

y=kxy = \frac{k}{x}

y=axy = a^x

💡Examples

Problem 1:

Sketch the curve y=x2−5x+6y = x^2 - 5x + 6 by finding the intercepts and the turning point.

Solution:

  1. Find the yy-intercept: Set x=0x = 0: y=02−5(0)+6=6y = 0^2 - 5(0) + 6 = 6 So, the yy-intercept is (0,6)(0, 6).

  2. Find the xx-intercepts: Set y=0y = 0: x2−5x+6=0x^2 - 5x + 6 = 0 (x−2)(x−3)=0(x - 2)(x - 3) = 0 x=2x = 2 or x=3x = 3 So, the xx-intercepts are (2,0)(2, 0) and (3,0)(3, 0).

  3. Find the Turning Point (Vertex): The xx-coordinate is x=−b2a=−−52(1)=2.5x = -\frac{b}{2a} = -\frac{-5}{2(1)} = 2.5 Substitute x=2.5x = 2.5 into the equation: y=(2.5)2−5(2.5)+6y = (2.5)^2 - 5(2.5) + 6 y=6.25−12.5+6=−0.25y = 6.25 - 12.5 + 6 = -0.25 The turning point is (2.5,−0.25)(2.5, -0.25).

  4. Sketching: Draw a 'U' shaped parabola passing through (0,6)(0, 6), (2,0)(2, 0), and (3,0)(3, 0) with the lowest point at (2.5,−0.25)(2.5, -0.25).

Explanation:

To sketch a quadratic, you need the orientation (positive x2x^2 means 'U' shape), the points where it crosses the axes, and the coordinate of the peak or valley.

Problem 2:

Identify the features of the reciprocal graph y=4xy = \frac{4}{x} for a sketch.

Solution:

  1. Asymptotes: As xx gets very large, yy approaches 00. The line y=0y = 0 (x-axis) is a horizontal asymptote. As xx approaches 00, yy becomes infinitely large or small. The line x=0x = 0 (y-axis) is a vertical asymptote.

  2. Key points: If x=1,y=4x = 1, y = 4 If x=4,y=1x = 4, y = 1 If x=−1,y=−4x = -1, y = -4 If x=−4,y=−1x = -4, y = -1

  3. Sketching: The graph exists in the 1st and 3rd quadrants. It consists of two separate curves that never touch the axes.

Explanation:

Reciprocal graphs are discontinuous at x=0x = 0 and are characterized by their asymptotic behavior towards the axes.