Some equations are hard to solve by algebra, but easy to solve with a graph: draw the right line, find where it crosses the curve, and read down to the \(x\)-axis.
The solutions of \(\text{curve} = \text{line}\) are the \(x\)-coordinates of the points where the two graphs cross. To solve a new equation with a curve you already have, rearrange it so one side is exactly the curve's equation; the other side is the line to draw.
Fig. 1: y = x² − 2x − 2 crosses the x-axis at about −0.7 and 2.7, and meets the line y = x + 2 at (−1, 1) and (4, 6)
Section 01
Roots are on the x-axis.
Open the solving tab in the lab. Pick an equation, then choose the line you think will solve it. Start with the easiest case, where the line is \(y = 0\), the \(x\)-axis, and read off where the curve crosses it.
A root of \(f(x) = 0\) is an \(x\)-value that makes the expression zero: on the graph it is where the curve crosses (or touches) the \(x\)-axis.
Reading accuracy: read to the nearest small square, usually to 1 decimal place. Exact values are not expected from a graph.
How many? Count the crossings: a U-shaped parabola can give 2, 1 or 0 roots.
Worked example Roots from the graph
Use Fig. 1 to solve \(x^2 - 2x - 2 = 0\).
The left side is exactly the curve \(y = x^2 - 2x - 2\), and the right side is 0, so the line to use is \(y = 0\): the \(x\)-axis.
The curve crosses the \(x\)-axis twice.
Read down at each crossing: \(x \approx -0.7\) and \(x \approx 2.7\).
Use the graph of \(y = x^2 - 2x - 2\) to solve \(x^2 - 2x - 2 = 3\).
The left side is the curve, so draw the horizontal line \(y = 3\).
Find where \(y = 3\) meets the curve: two points.
Read down to the \(x\)-axis: about \(-1.4\) and \(3.4\).
Answer: \(x \approx -1.4\) or \(x \approx 3.4\)
Examiner trap
Give the \(x\)-values, not the \(y\)-value. For \(x^2 - 2x - 2 = 3\), the answer is NOT \(y = 3\); it is the two \(x\)-coordinates where the line \(y = 3\) cuts the curve. And give BOTH solutions.
Section 02
A line meets a curve.
When a straight line and a curve are drawn on the same grid, each crossing point satisfies both equations at once, so its \(x\)-coordinate solves \(\text{curve} = \text{line}\).
Worked example Reading intersections
In Fig. 1 the line \(y = x + 2\) is drawn on the graph of \(y = x^2 - 2x - 2\). Write down the equation solved by the crossing points, and solve it.
At the crossings the \(y\)-values are equal: \(x^2 - 2x - 2 = x + 2\).
Tidy up: \(x^2 - 3x - 4 = 0\).
The graphs cross at \((-1, 1)\) and \((4, 6)\), so \(x = -1\) or \(x = 4\).
Check by factorising: \((x + 1)(x - 4) = 0\) gives the same values.
The graph of \(y = x^2 - 2x - 2\) is drawn. Find the equation of the straight line you should draw to solve \(x^2 - 3x - 1 = 0\).
Aim: make the left side look exactly like the curve, \(x^2 - 2x - 2\).
To turn \(-3x\) into \(-2x\), add \(x\). To turn \(-1\) into \(-2\), subtract 1.
Do the same to both sides: \(x^2 - 3x - 1 + x - 1 = 0 + x - 1\).
So \(x^2 - 2x - 2 = x - 1\). The right side is the line.
Answer: Draw \(y = x - 1\). It meets the curve at \(x \approx -0.3\) and \(x \approx 3.3\).
Your turn
Using the same curve, how many solutions does \(x^2 - 2x - 2 = k\) have when (a) \(k = 1\) (b) \(k = -3\) (c) \(k = -5\)?
The lowest point of the curve is \((1, -3)\). (a) The line \(y = 1\) is above it: 2 solutions. (b) \(y = -3\) just touches the bottom: 1 solution. (c) \(y = -5\) is below the curve: no solutions.
Examiner trap
When you rearrange, change BOTH sides by the same amount. Students often move terms across and write \(y = 3x + 1\) for this question, but the line must come from keeping the equation balanced: \(y = x - 1\).
Stage 1 · Learn
Check your understanding
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Quick check+10 XP
The graph of \(y = f(x)\) crosses the \(x\)-axis at \(x = -1\) and \(x = 5\). What are the solutions of \(f(x) = 0\)?
Quick check+10 XP
You have the graph of \(y = x^2 - 2x - 2\). Which line do you draw to solve \(x^2 - 2x - 2 = 3\)?
Quick check+10 XP
A line and a curve cross at \((-1, 1)\) and \((4, 6)\). What are the solutions of \(\text{curve} = \text{line}\)?
Quick check+10 XP
The lowest point of \(y = x^2 - 2x - 2\) is \((1, -3)\). How many solutions has \(x^2 - 2x - 2 = -5\)?
Section 03
Cubic and reciprocal curves.
Extended. The method is identical for any curve: make one side of the equation match the curve that is drawn, and draw the other side as a line.
Worked example A reciprocal and a line (Extended)
Show that the \(x\)-coordinates of the points where \(y = \dfrac{6}{x}\) meets \(y = x + 1\) satisfy \(x^2 + x - 6 = 0\), and find them.
Equal \(y\)-values: \(\dfrac{6}{x} = x + 1\).
Multiply both sides by \(x\): \(6 = x^2 + x\).
Subtract 6: \(x^2 + x - 6 = 0\).
Factorise: \((x + 3)(x - 2) = 0\), so the graphs meet at \(x = -3\) and \(x = 2\).
Answer: \(x = -3\) or \(x = 2\), at the points \((-3, -2)\) and \((2, 3)\)
Worked example Which line for a cubic? (Extended)
The graph of \(y = x^3 - 3x + 1\) is drawn. Find the line to draw to solve \(x^3 - 5x + 2 = 0\).
Make the left side \(x^3 - 3x + 1\): add \(2x\) and subtract 1.
\(x^3 - 5x + 2 + 2x - 1 = 2x - 1\).
So \(x^3 - 3x + 1 = 2x - 1\).
Answer: Draw \(y = 2x - 1\). (The solutions are \(x = 2\) and \(x \approx -2.4\) and \(0.4\).)
Examiner trap
Only read solutions inside the drawn range. If a line leaves the grid before meeting the curve again, say there are no more solutions in that range rather than guessing.
Stage 2 · Exam
Exam-style questions
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The graph of \(y = x^2 - 2x - 2\) is drawn. Which straight line should be drawn to solve \(x^2 - 3x - 1 = 0\)?
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(Extended) The graph of \(y = x^3 - 3x + 1\) is drawn. Which straight line should be drawn to solve \(x^3 - 5x + 2 = 0\)?
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The line \(y = x + 2\) meets the curve \(y = x^2 - 2x - 2\) at \(A\) and \(B\). Find the coordinates of \(A\) and \(B\).
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The lowest point of \(y = x^2 - 2x - 2\) is \((1, -3)\). For which values of \(k\) does \(x^2 - 2x - 2 = k\) have two solutions?
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