Plotting quadratic graphs Wobbly curve when the points chosen are too far apart Feathering of the line when the curve is drawn in small sections Marks can be lost in the following ways… Line drawn through a point that is clearly incorrect Flat base to the curve where it should curve smoothly

Possible exam question: Plot the graphs of these two functions, and state the co-ordinates where they cross: y = x2 + 2 y = 2x2 – 4x + 2 Co-ordinates: (0,2)

Plot these graphs and find the points where they cross: y = 2x2 – 1 y = -x2 + 2 X -3 -2 -1 1 2 3 y

A B C D Significant points on your graph A and B: known as the roots of the equation, when y = 0. Found by drawing the graph and seeing where it crosses the x axis. C: the y-axis intercept. Found by putting x = 0. D: the vertex of the graph. The x-coordinate is always half way between the roots.

Copy and complete the table below to plot the graphs of: y = x2 – 4 y = x2 – 9 y = x2 + 4x y = x2 – 6x Use the graphs to find the roots of these equations when y = 0. Once you have done one, see if you can predict what the roots of the others will be. X -4 -3 -2 -1 1 2 3 4 Y

Solving equations using other graphs Possible exam question: The diagram below shows the curve y = x2 + 3x - 2 By drawing a suitable straight line, solve the equation: x2 + 3x - 1 = 0 Given Graph: y = x2 + 3x - 2 New Equation: 0 = x2 + 3x - 1 Subtract: y = - 1 Draw y = -1 Solutions: x = 0.3, -3.3

Your turn….remember, follow the steps just like the last question. Example: The diagram below shows the curve y = x2 + 3x - 2 By drawing a suitable straight line, solve the equation: x2 + 2x - 3 = 0 Given Graph: y = x2 + 3x - 2 New Equation: 0 = x2 + 2x - 3 Subtract: y = + x + 1 Draw y = x + 1 Solutions: x = 1, -3