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Algebra - Functions and graphs

Grade 7Cambridge (IGCSE)

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

🔑Concepts

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A function f(x)f(x) is a rule that assigns each input value xx to exactly one output value yy. In a coordinate plane, the graph of a linear function is a straight line, while non-linear functions like quadratics form curves like parabolas.

A straight line graph of y = x + 1 on a coordinate plane.
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The gradient (mm) represents the steepness of a line. A positive gradient slopes upwards from left to right, while a negative gradient slopes downwards. The yy-intercept (cc) is the point where the line crosses the vertical axis.

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Parallel lines have the same gradient. If two lines are parallel, their equations will have the same mm value in the form y=mx+cy = mx + c.

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Vertical lines are written as x=kx = k and horizontal lines as y=ky = k, where kk is a constant. A vertical line has an undefined gradient, while a horizontal line has a gradient of 00.

📐Formulae

y=mx+cy = mx + c (Equation of a straight line)

m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} (Gradient formula)

y−y1=m(x−x1)y - y_1 = m(x - x_1) (Point-gradient form)

M=(x1+x22,y1+y22)M = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}) (Midpoint of a line segment)

💡Examples

Problem 1:

Complete the table of values for the function y=3x−2y = 3x - 2 for xx values −1,0,1,2-1, 0, 1, 2.

Solution:

For x=−1:y=3(−1)−2=−5x = -1: y = 3(-1) - 2 = -5; For x=0:y=3(0)−2=−2x = 0: y = 3(0) - 2 = -2; For x=1:y=3(1)−2=1x = 1: y = 3(1) - 2 = 1; For x=2:y=3(2)−2=4x = 2: y = 3(2) - 2 = 4. The coordinates are (−1,−5),(0,−2),(1,1),(2,4)(-1, -5), (0, -2), (1, 1), (2, 4).

Explanation:

Substitute each given xx value into the equation y=3x−2y = 3x - 2 to find the corresponding yy value, then pair them as coordinates.

Problem 2:

Find the gradient of the line passing through the points A(2,5)A(2, 5) and B(4,13)B(4, 13).

Solution:

m=13−54−2=82=4m = \frac{13 - 5}{4 - 2} = \frac{8}{2} = 4.

Explanation:

Apply the gradient formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} by identifying x1=2,y1=5x_1=2, y_1=5 and x2=4,y2=13x_2=4, y_2=13.

Problem 3:

Identify the gradient and y-intercept of the line with the equation 2y=6x+42y = 6x + 4.

Solution:

Divide by 2 to get y=3x+2y = 3x + 2. Gradient (mm) = 3, Y-intercept (cc) = 2.

Explanation:

To find mm and cc, the equation must first be rearranged into the standard form y=mx+cy = mx + c.

Problem 4:

What is the equation of the horizontal line that passes through the point (5,−3)(5, -3)?

Solution:

y=−3y = -3

Explanation:

A horizontal line has the same y-coordinate for every point on the line. Since it passes through (5,−3)(5, -3), the y-value is always −3-3.

Problem 5:

Determine the equation of the line shown in the diagram that passes through (0,2)(0, 2) and (4,0)(4, 0).

Line passing through (0,2) and (4,0).

Solution:

m=0−24−0=−24=−0.5m = \frac{0 - 2}{4 - 0} = \frac{-2}{4} = -0.5 The y-intercept cc is 22 (where x=0x=0). Equation: y=−0.5x+2y = -0.5x + 2

Explanation:

First, calculate the gradient using the two points provided. Since the line crosses the y-axis at 2, c=2c = 2. Substitute mm and cc into y=mx+cy = mx + c.

Problem 6:

Identify the coordinates of the vertex (turning point) for the quadratic function y=x2−4y = x^2 - 4 as shown in the graph.

Graph of the parabola y = x^2 - 4.

Solution:

The vertex is at (0,−4)(0, -4).

Explanation:

The vertex of a parabola y=ax2+cy = ax^2 + c is the highest or lowest point. For y=x2−4y = x^2 - 4, the lowest point occurs when x=0x = 0, giving y=−4y = -4.

Functions and graphs Grade 7 Notes & Examples