1. 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

2. 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)

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

4. 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.

5. 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

6. 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 =
Draw y = -1
Solutions: x = 0.3, -3.3

7. 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