# StudySquare

# Solving equations graphically for OCR A-level Maths

This page covers the following topics:

1. Equations of a straight line

2. Quadratic equations from graphs

3. Solutions of equations from graphs

4. Graphs of complex equations

A straight line graph has an equation of y = mx + c, where m is the gradient of the graph and c is the y-intercept of the graph, ie. the point at which the line crosses the y-axis. The gradient of the line can be calculated using: m = change in y/change in x. Another equation that can be used to find the equation of a straight line is y − y₁ = m(x − x₁), where m is the gradient of the graph and the graph passes through the point (x₁, y₁).

A quadratic equation is written in the form y = ax² + bx + c. To find the equation of a quadratic from its graph, the values of a, b and c must be calculated. To do so, three points on the graph must be known. These can be substituted in the general form of a quadratic, which would give 3 equations that can be solved simultaneously to find the values of a, b and c.

Systems of simultaneous equations can be solved by graphing: the solutions to the system will be the points of intersection on their graphs.

Graphs that involve negative quadratic equations and cubic equations can be solved just as linear and quadratic equations. The typical shapes of a negative quadratic and cubic graphs are given in the diagram.

# 1

Find the equation of the line with gradient 7 that passes through the point of intersection of the lines 2x − y = 19 and 3y + 4x + 57 = 0.

Rearranging the first equation, y = 2x − 19.

Substituting this into the second equation, 3(2x − 19) + 4x + 57 = 0.

By expansion 6x − 57 + 4x + 57 = 0, thus 10x = 0 and x = 0.

Substituting gives y = 2(0) − 19 = −19.

Therefore, the point of intersection is (0, −19), which is the y-intercept of the line.

Using y = mx + c, the gradient of 7 and the y-intercept of −19, the equation of the line is y = 7x − 19.

y = 7x − 19

# 2

The graph of y = x² − 5x − 14 and an unknown linear equation are given in the diagram. Find the coordinates of the point of intersection.

The graph of the linear equation intersects the graph of the quadratic at its negative root.

The roots of the quadratic are at y = 0, therefore 0 = x² − 5x − 14 must be solved to find the roots.

Factorising gives 0 = (x + 2)(x − 7), and equating each factor to 0 gives x + 2 = 0 or x − 7 = 0.

Therefore, x = −2 or x = 7, so the negative root of the quadratic is at x = −2.

Thus, the coordinates of the point of intersection are (−2, 0).

(−2, 0)

# 3

Using the points given on the diagram, determine the equation of the graph.

The three points on the graph are (0, 3), (4, 5) and (−4, 5).

Substituting x = 0 and y = 3 into the general form of a quadratic, 3 = a(0)² + b(0) + c, so c = 3.

Substituting x = 4, y = 5 and c = 3, 5 = a(4)² + b(4) + 3, so 2 = 16a + 4b.

Substituting x = −4, y = 5 and c = 3, 5 = a(−4)² + b(−4) + 3, so 2 = 16a − 4b.

Adding these two equations gives 4 = 32a, so a = 1/8.

Substituting this into the first equation, 2 = 16(1/8) + 4b, so 4b = 0 and b = 0.

Therefore, the equation of the quadratic is y = (1/8)x² + 3.

y = (1/8)x² + 3

# 4

The graphs of two equations are drawn on the same set of axes. Given that the two graphs intersect twice, state the number of solutions for this pair of simultaneous equations and explain your answer.

When solving simultaneous equations graphically, the points of intersection are the solutions to the simultaneous equations. Since the graphs of these two equations intersect twice, the system of simultaneous equations has two solutions.

2

# 5

The solutions to a quadratic equation are x = −1 and x = 11. The y-intercept of the graph of the quadratic is −154. Find the equation of the quadratic graph.

Since the solutions of the quadratic equation are x = −1 and x = 11, the graph of the quadratic passes through the points (−1, 0) and (11, 0).

Since the y-intercept of the graph is −154, the graph passes through the point (0, −154).

Substituting x = 0 and y = −154 into the general form of a quadratic, −154 = a(0)² + b(0) + c, so c = −154.

Substituting x = −1, y = 0 and c = −154, 0 = a(−1)² + b(−1) − 154, so 154 = a − b.

Substituting x = 11, y = 0 and c = −154, 0 = a(11)² + b(11) − 154, so 154 = 121a + 11b.

Rearranging the first equation gives a = 154 + b.

Substituting this into the second equation gives 154 = 121(154 + b) + 11b.

Simplifying gives 154 = 18634 + 132b, so b = −140.

Substituting this into the equation for a gives a = 154 − 140 = 14.

Therefore, the quadratic equation is y = 14x² − 140x − 154.

y = 14x² − 140x − 154

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