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To find the equation of straight line graphs students need to calculate the gradient using two pairs of coordinates and the intercept which is the y value of where it crosses the vertical axis. Examiner’s reports of past exam questions show students are more able to find the intercept of a straight line than they are at calculating the gradient. This lesson on interpreting straight line graphs is available here.

The starter recaps the learning in the previous lesson on using graphs to solve equations by interpreting the x and y values of a line. By adding the x and y values together students can see the equation of the line is x + y = 5.

As an extension I ask the class to imagine the line extended beyond the grid and to show me other coordinate pairs the line would pass through. We have a short discussion about coordinate pairs with decimal numbers and move on to the main teaching part.

To introduce the main part of the lesson we discuss how the steepness (gradient, M) and position (intercept, C) of where the line crosses the y axis define how the graph looks on a grid. Equations of linear graphs can be written in the form y = mx + c.

The gradient (M) can be calculated using two coordinates that lie along a line. The gradient is the change in vertical distance divided by the change in horizontal distance. This gives a measure of the steepness of the line.

The intercept (C) is the point where the line crosses the y axis. The intercept can also be defined algebraically as the value of y when x = 0.

I work through the first two examples on the second slide with the class and encourage the students to attempt the third question on their mini-whiteboards. If more practice is needed I use the Interactive Excel File to generate further questions. When they are ready students work through the questions on the third slide independently. All the answers are provided in the lesson plan.

The plenary extends the learning because the line now has a fractional gradient, negative intercept and students need to choose their own coordinates to use.

Rather than calculating the gradient they are asked to explain why it is 2/3. To do this they simply use the method as before leaving the answer as a fraction. I encourage students to use two sets of coordinates, so they can check their working. This also consolidates their method. The equation of the straight line is y = (2/3)x – 1.

The next lesson in the Functions, Graphs and Equations unit goes onto plotting quadratic graphs from a table of results where students explore the properties of parabolas.

March 10, 2019

When calculating the volume of a pyramid we can substitute the values of the length, width and perpendicular height into the formula V = 1/3 lwh. In my experience this is often provided for the students with little explanation as to why a volume of a pyramid is exactly one third the volume of a […]

March 4, 2019

When teaching solving 3D problems using trigonometry we begin the lesson with a recap of Pythagoras’ Theorem and the three trigonometric ratios. We do this by matching the ratio and equations to the respective right-angled triangle. Students are encouraged to work in pairs and to show the diagrams as part of the working out on […]

January 29, 2019

When I teach rounding to a significant figure, I ask the class to discuss in pairs or small groups a definition for the word significant. It is a word that all the students have heard before but not all are able to define. After 2 or 3 minutes of conversation I ask the students to […]