Sec 3 E-Math · Graphs of functions

Solving equations graphically questions with answers

Practice questions on solving equations graphically, from the basics up to exam level, each with a full worked solution. Try them on paper first, then open the solution to check your working.

Worked example

The diagram shows the graph of $y = -\dfrac{4}{x}$ and the line $y = 2$ . Use the graph to solve the equation $-\dfrac{4}{x} = 2$ .

Step 1 of 3

Each side of the equation is a graph

Left-hand side:

y = -\dfrac{4}{x}

, the curveRight-hand side:

y = 2

, the horizontal line

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Questions

Question 1Curve and a horizontal line
The diagram shows the graph of $y = x^{2} - 2x$ and the line $y = 0$ . Use the graph to solve the equation $x^{2} - 2x = 0$ .
Show the worked solution
1. Step 1: Each side of the equation is a graphLeft-hand side: $y = x^{2} - 2x$ , the curveRight-hand side: $y = 0$ , the horizontal line
2. Step 2: The solution is where the LHS graph meets the RHS graphWhere the curve and the line meet, both sides have the same $y$ -value for the same $x$ , so the equation is true there.Number of points where they meet: $2$
3. Step 3: Read the x-coordinate of each one $x = 0$ or $x = 2$
Question 2Curve and a sloping line
The diagram shows the graph of $y = x^{3} - 3x$ and the line $y = -x + 4$ . Use the graph to solve the equation $x^{3} - 3x = -x + 4$ .
Show the worked solution
1. Step 1: Each side of the equation is a graphLeft-hand side: $y = x^{3} - 3x$ , the curveRight-hand side: $y = -x + 4$ , the line
2. Step 2: The solution is where the LHS graph meets the RHS graphWhere the curve and the line meet, both sides have the same $y$ -value for the same $x$ , so the equation is true there.Number of points where they meet: $1$
3. Step 3: Read the x-coordinate of each one $x = 2$
Question 3Which line to draw: quadratic
The graph of $y = x^{2} + 2x - 2$ has been drawn. Find the equation of the straight line that must be drawn on the same axes to solve the equation $x^{2} - 1 = 0$ .
Show the worked solution
1. Step 1: Rearrange so the drawn curve is the left-hand side $x^{2} - 1 = 0$ $x^{2} + 2x - 2 = 2x - 1$
2. Step 2: The right-hand side is the line to drawThe solutions are where the LHS graph $y = x^{2} + 2x - 2$ meets the RHS graph, so draw $y = 2x - 1$ .Then read the $x$ -coordinates where the curve and the line meet.
3. Step 3: Check: it must be a straight line y = mx + cGradient $2$ , $y$ -intercept $-1$ .
Question 4Which line to draw: cubic, reciprocal
The graph of $y = x^{3}$ has been drawn. Find the equation of the straight line that must be drawn on the same axes to solve the equation $x^{3} + x - 2 = 0$ .
Show the worked solution
1. Step 1: Rearrange so the drawn curve is the left-hand side $x^{3} + x - 2 = 0$ $x^{3} = -x + 2$
2. Step 2: The right-hand side is the line to drawThe solutions are where the LHS graph $y = x^{3}$ meets the RHS graph, so draw $y = -x + 2$ .Then read the $x$ -coordinates where the curve and the line meet.
3. Step 3: Check: it must be a straight line y = mx + cGradient $-1$ , $y$ -intercept $2$ .
Question 5Read to 1 decimal place
The diagram shows the graph of $y = -\dfrac{4}{x}$ and the line $y = -x - 2$ . Use the graph to solve the equation $-\dfrac{4}{x} = -x - 2$ , giving your answers correct to 1 decimal place.
Show the worked solution
1. Step 1: Each side of the equation is a graphLeft-hand side: $y = -\dfrac{4}{x}$ , the curveRight-hand side: $y = -x - 2$ , the line
2. Step 2: The solution is where the LHS graph meets the RHS graphWhere the curve and the line meet, both sides have the same $y$ -value for the same $x$ , so the equation is true there.Number of points where they meet: $2$
3. Step 3: Read the x-coordinate of each one to 1 decimal place $x = -3.2$ or $x = 1.2$ (1 d.p.)
Question 6Find the line drawn, then read the solutions
The diagram shows the graph of $y = -x^{3} + 3x + 1$ , and the straight line that must be drawn on the same axes to solve the equation $-x^{3} + 2x + 4 = 0$ . Use the graph to solve this equation.
Show the worked solution
1. Step 1: Rearrange so the drawn curve is the left-hand side $-x^{3} + 2x + 4 = 0$ $-x^{3} + 3x + 1 = x - 3$
2. Step 2: The line drawn is y = the right-hand side: the solutions are where it meets the curveThe line drawn is $y = x - 3$ .Number of points where they meet: $1$
3. Step 3: Read the x-coordinate of each one $x = 2$
Question 7A range of x from the graph
The diagram shows the graph of $y = x^{2} - 2x$ and the line $y = -2x + 4$ . Use the graph to find the range of values of $x$ for which $x^{2} - 2x < -2x + 4$ .
Show the worked solution
1. Step 1: Where do the curve and the line meet? $x = -2$ or $x = 2$
2. Step 2: Where is the curve below the line?The curve is below the line between the two crossings.
3. Step 3: Write the range of values of x $-2 < x < 2$
Question 8How many solutions? Values of k
The diagram shows the graph of $y = x^{2} - 2x - 3$ . Use the graph to find the number of solutions of the equation $x^{2} - 2x - 3 = -6$ .
Show the worked solution
1. Step 1: Read the turning point: the minimum point of the curve $y$ -coordinate: $-4$
2. Step 2: Draw the line y = -6: where is it? $y = -6$ is below the minimum.
3. Step 3: Count the points where the line meets the curveNumber of solutions: $0$

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Worked example

The diagram shows the graph of

y = -\dfrac{4}{x}

and the line

y = 2

. Use the graph to solve the equation

-\dfrac{4}{x} = 2

Step 1 of 3

Each side of the equation is a graph

Left-hand side:

y = -\dfrac{4}{x}

, the curveRight-hand side:

y = 2

, the horizontal line

Questions

Question 1Curve and a horizontal line

The diagram shows the graph of $y = x^{2} - 2x$ and the line $y = 0$ . Use the graph to solve the equation $x^{2} - 2x = 0$ .

Show the worked solution

Step 1: Each side of the equation is a graphLeft-hand side: $y = x^{2} - 2x$ , the curveRight-hand side: $y = 0$ , the horizontal line
Step 2: The solution is where the LHS graph meets the RHS graphWhere the curve and the line meet, both sides have the same $y$ -value for the same $x$ , so the equation is true there.Number of points where they meet: $2$
Step 3: Read the x-coordinate of each one $x = 0$ or $x = 2$

Question 2Curve and a sloping line

The diagram shows the graph of $y = x^{3} - 3x$ and the line $y = -x + 4$ . Use the graph to solve the equation $x^{3} - 3x = -x + 4$ .

Show the worked solution

Step 1: Each side of the equation is a graphLeft-hand side: $y = x^{3} - 3x$ , the curveRight-hand side: $y = -x + 4$ , the line
Step 2: The solution is where the LHS graph meets the RHS graphWhere the curve and the line meet, both sides have the same $y$ -value for the same $x$ , so the equation is true there.Number of points where they meet: $1$
Step 3: Read the x-coordinate of each one $x = 2$

Question 3Which line to draw: quadratic

The graph of $y = x^{2} + 2x - 2$ has been drawn. Find the equation of the straight line that must be drawn on the same axes to solve the equation $x^{2} - 1 = 0$ .

Show the worked solution

Step 1: Rearrange so the drawn curve is the left-hand side $x^{2} - 1 = 0$ $x^{2} + 2x - 2 = 2x - 1$
Step 2: The right-hand side is the line to drawThe solutions are where the LHS graph $y = x^{2} + 2x - 2$ meets the RHS graph, so draw $y = 2x - 1$ .Then read the $x$ -coordinates where the curve and the line meet.
Step 3: Check: it must be a straight line y = mx + cGradient $2$ , $y$ -intercept $-1$ .

Question 4Which line to draw: cubic, reciprocal

The graph of $y = x^{3}$ has been drawn. Find the equation of the straight line that must be drawn on the same axes to solve the equation $x^{3} + x - 2 = 0$ .

Show the worked solution

Step 1: Rearrange so the drawn curve is the left-hand side $x^{3} + x - 2 = 0$ $x^{3} = -x + 2$
Step 2: The right-hand side is the line to drawThe solutions are where the LHS graph $y = x^{3}$ meets the RHS graph, so draw $y = -x + 2$ .Then read the $x$ -coordinates where the curve and the line meet.
Step 3: Check: it must be a straight line y = mx + cGradient $-1$ , $y$ -intercept $2$ .

Question 5Read to 1 decimal place

The diagram shows the graph of $y = -\dfrac{4}{x}$ and the line $y = -x - 2$ . Use the graph to solve the equation $-\dfrac{4}{x} = -x - 2$ , giving your answers correct to 1 decimal place.

Show the worked solution

Step 1: Each side of the equation is a graphLeft-hand side: $y = -\dfrac{4}{x}$ , the curveRight-hand side: $y = -x - 2$ , the line
Step 2: The solution is where the LHS graph meets the RHS graphWhere the curve and the line meet, both sides have the same $y$ -value for the same $x$ , so the equation is true there.Number of points where they meet: $2$
Step 3: Read the x-coordinate of each one to 1 decimal place $x = -3.2$ or $x = 1.2$ (1 d.p.)

Question 6Find the line drawn, then read the solutions

The diagram shows the graph of $y = -x^{3} + 3x + 1$ , and the straight line that must be drawn on the same axes to solve the equation $-x^{3} + 2x + 4 = 0$ . Use the graph to solve this equation.

Show the worked solution

Step 1: Rearrange so the drawn curve is the left-hand side $-x^{3} + 2x + 4 = 0$ $-x^{3} + 3x + 1 = x - 3$
Step 2: The line drawn is y = the right-hand side: the solutions are where it meets the curveThe line drawn is $y = x - 3$ .Number of points where they meet: $1$
Step 3: Read the x-coordinate of each one $x = 2$

Question 7A range of x from the graph

The diagram shows the graph of $y = x^{2} - 2x$ and the line $y = -2x + 4$ . Use the graph to find the range of values of $x$ for which $x^{2} - 2x < -2x + 4$ .

Show the worked solution

Step 1: Where do the curve and the line meet? $x = -2$ or $x = 2$
Step 2: Where is the curve below the line?The curve is below the line between the two crossings.
Step 3: Write the range of values of x $-2 < x < 2$

Question 8How many solutions? Values of k

The diagram shows the graph of $y = x^{2} - 2x - 3$ . Use the graph to find the number of solutions of the equation $x^{2} - 2x - 3 = -6$ .

Show the worked solution

Step 1: Read the turning point: the minimum point of the curve $y$ -coordinate: $-4$
Step 2: Draw the line y = -6: where is it? $y = -6$ is below the minimum.
Step 3: Count the points where the line meets the curveNumber of solutions: $0$