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Using Graphs (Cambridge O Level Maths)
Revision Note
Solving Equations Using Graphs
How do we use graphs to solve equations?
- Solutions are always read off the x-axis
- Solutions of f(x) = 0 are where the graph of y = f(x) crosses the x-axis
- If asked to use the graph of y = f(x) to solve a different equation (the question will say something like “by drawing a suitable straight line”) then:
- Rearrange the equation to be solved into f(x) = mx + c and draw the line y = mx + c
- Solutions are the x-coordinates of where the line (y = mx + c) crosses the curve (y = f(x))
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E.g. if given the curve for y = x3 + 2x2 + 1 and asked to solve x3 + 2x2 − x − 1 = 0, then;
- rearrange x3 + 2x2 − x − 1 = 0 to x3 + 2x2 + 1 = x + 2
- draw the line y = x + 2 on the curve y = x3 + 2x2 + 1
- read the x-values of where the line and the curve cross (in this case there would be 3 solutions, approximately x = -2.2, x = -0.6 and x = 0.8);
- Note that solutions may also be called roots
How do we use graphs to solve linear simultaneous equations?
- Plot both equations on the same set of axes using straight line graphs y = mx + c
- Find where the lines intersect (cross)
- The x and y solutions to the simultaneous equations are the x and y coordinates of the point of intersection
- e.g. to solve 2x - y = 3 and 3x + y = 4 simultaneously, first plot them both (see graph)
- find the point of intersection, (2, 1)
- the solution is x = 2 and y = 1
How do we use graphs to solve simultaneous equations where one is quadratic?
- e.g. to solve y = x2 + 4x − 12 and y = 1 simultaneously, first plot them both (see graph)
- find the two points of intersection (by reading off your scale), (-6.1 , 1) and (-2.1, 1) to 1 decimal place
- the solutions from the graph are approximately x = -6.1 and y = 1 and x = 2.1 and y = 1
- note their are two pairs of x, y solutions
- to find exact solutions, use algebra
Examiner Tip
- If solving an equation, give the x values only as your final answer
- If solving a pair of linear simultaneous equations give an x and a y value as your final answer
- If solving a pair of simultaneous equations where one is linear and one is quadratic, give two pairs of x and y values as your final answer
Worked example
The graph of is shown below.
Use the graph to estimate the solutions of the equation . Give your answers to 1 decimal place.
We are given a different equation to the one plotted so we must rearrange it to (where is the plotted graph)
Now plot on the graph- this is the solid red line on the graph below
The solutions are the coordinates of where the curve and the straight line cross so
Finding Gradients of Tangents
What is the gradient of a graph?
- The gradient of a graph at any point is equal to the gradient of the tangent to the curve at that point
- Remember that a tangent is a line that just touches a curve (and doesn’t cross it)
How do I estimate the gradient under a graph?
- To find an estimate for the gradient:
- Draw a tangent to the curve
- Find the gradient of the tangent using Gradient = RISE ÷ RUN
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- In the example above, the gradient at x = 4 would be
- It is an estimate because the tangent has been drawn by eye and is not exact
- (To find the exact gradient we would need to use differentiation)
What does the gradient represent?
- In a y-x graph, the gradient represents the rate of change of y against x
- This has many practical applications, for example;
- in a distance-time graph, the gradient (rate of change of distance against time) is the speed
- in a speed-time graph, the gradient (rate of change of speed against time) is the acceleration
Examiner Tip
- This is particularly useful when working with Speed-Time and Distance-Time graphs if they are curves and not straight lines
Worked example
The graph below shows for
Find an estimate of the gradient of the curve at the point where
Draw a tangent to the curve at the point where x = 0.5.
Find suitable, easy to read coordinates and draw a right-angled triangle between them.
Find the difference in the y coordinates (rise) and the difference in the x coordinates (run).
Divide the difference in y (rise) by the difference in x.
Estimate of gradient = 0.6
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