Finding Approximate Solutions Using Graphs Revision Notes | GCSE AQA Higher Maths

1. Why Graphs Can Solve Equations

Some equations refuse to factorise neatly. Others involve a mix of $x^2$ and $x$ terms that make algebraic methods messy. Graphs give you a visual shortcut: wherever two curves cross, the $x$-value at that crossing is a solution to the equation you care about.

Think of it this way. If you plot $y = x^2 - 3x - 1$ and $y = 2$, the points where the parabola meets the horizontal line are exactly the $x$-values where $x^2 - 3x - 1 = 2$. Reading those $x$-values off the graph gives you approximate solutions without any algebra at all.

The word approximate is key here. When you read a value from a graph, you are estimating to the nearest small division on the axis. GCSE Higher exam questions will ask you to give answers to one decimal place, which is the level of precision a well-drawn graph supports.

2. Reading Solutions Directly from a Graph

The most straightforward type of question gives you a graph already drawn and asks you to use it to solve an equation. Your job is to identify which two graphs need to intersect, draw any missing line if needed, and read the $x$-coordinates of the crossing points.

The graph above shows $y = x^2 - 3x - 1$ (a parabola) and $y = 2$ (a horizontal line). The parabola crosses the line at two points. Reading the $x$-coordinates of those intersections gives the approximate solutions to $x^2 - 3x - 1 = 2$, which are approximately $x \approx -0.6$ and $x \approx 3.6$.

Worked example: Reading Approximate Solutions from a Given Graph

The graph of $y = x^2 - 3x - 1$ is drawn. Use the graph to find approximate solutions to $x^2 - 3x - 1 = 2$.

1. $x^2 - 3x - 1 = 2$ — The left-hand side matches the curve already drawn, so we need to find where the curve equals 2.
2. Draw $y = 2$ as a horizontal line across the grid. — The right-hand side is a constant, so the second graph is a straight horizontal line at height 2. [M1 for drawing the correct line]
3. The parabola crosses $y = 2$ at two points. Drop vertical lines from each crossing to the $x$-axis. — Each crossing point gives one solution. Reading carefully to the nearest 0.1 is expected.
4. $x \approx -0.6$ and $x \approx 3.6$ — [A1 for each correct value, accept answers in the range $\pm 0.1$ of the exact values] These are the approximate solutions.

Answer: $x \approx -0.6$ and $x \approx 3.6$

3. Rearranging to Use a Given Curve

Harder questions give you a curve — say $y = x^2 + x - 3$ — and ask you to solve a different equation using it. The trick is to rearrange that different equation until one side matches the curve exactly. Whatever is left on the other side becomes the line you draw.

This is where many students lose marks. They try to plot a brand-new curve instead of rearranging to use the one they already have. Rearranging is always faster and is what the examiner expects.

Worked example: Rearranging to Match a Given Curve

The graph of $y = x^2 + x - 3$ is drawn. Show that the equation $x^2 - x - 5 = 0$ can be solved using this graph, and state the line that should be drawn.

1. $x^2 - x - 5 = 0$ — This is the equation we want to solve. We need to manipulate it so that $x^2 + x - 3$ appears on one side.
2. $x^2 - x - 5 + 2x = 0 + 2x$ $x^2 + x - 5 = 2x$ — We are building towards $x^2 + x - 3$ on the left. Adding $2x$ gives us the $+x$ term we need.
3. $x^2 + x - 5 + 3 = 2x + 3 - 2$ Actually: $x^2 + x - 5 + 3 = 2x + 3$ $x^2 + x - 2 = 2x + 3$ — Let us redo this carefully. Starting from $x^2 + x - 5 = 2x$, add 3 to both sides: $x^2 + x - 5 + 3 = 2x + 3$, giving $x^2 + x - 2 = 2x + 3$. That does not yet match. Let us try a cleaner route.
4. Start: $x^2 - x - 5 = 0$ Add $2x$: $x^2 + x - 5 = 2x$ Add $2$: $x^2 + x - 3 = 2x + 2$ — Adding 2 to both sides gives $x^2 + x - 3$ on the left, which matches the curve exactly. [M1 for correct rearrangement]
5. Draw $y = 2x + 2$ on the same axes as $y = x^2 + x - 3$. — The intersections of $y = x^2 + x - 3$ and $y = 2x + 2$ give the solutions to $x^2 - x - 5 = 0$. [A1 for correct line equation]
6. The line $y = 2x + 2$ crosses the parabola at approximately $x \approx -1.8$ and $x \approx 2.8$. — [A1 for each value within $\pm 0.1$] These are the approximate solutions to $x^2 - x - 5 = 0$.

Answer: Draw $y = 2x + 2$; approximate solutions are $x \approx -1.8$ and $x \approx 2.8$.

4. Intersections of Two Non-Horizontal Graphs

Not every graphical solution involves a horizontal line. Sometimes you are given two curves — for example a quadratic and a straight line with a non-zero gradient — and asked to find where they intersect. The method is identical: the $x$-coordinates of the intersection points are the solutions to the equation you get by setting the two expressions equal.

The graph above shows $y = x^2 - x - 2$ and $y = x + 1$. Setting them equal gives $x^2 - x - 2 = x + 1$, which rearranges to $x^2 - 2x - 3 = 0$. The intersection points on the graph give the approximate solutions to this equation.

Worked example: Finding Solutions from Two Intersecting Graphs

The graphs of $y = x^2 - x - 2$ and $y = x + 1$ are drawn on the same axes. Use the graphs to find approximate solutions to $x^2 - 2x - 3 = 0$.

1. Set $x^2 - x - 2 = x + 1$ $x^2 - x - 2 - x - 1 = 0$ $x^2 - 2x - 3 = 0$ — This confirms that the crossing points of the two graphs do indeed solve $x^2 - 2x - 3 = 0$. [M1 for recognising the connection]
2. The parabola and the line cross at two points. Read the $x$-coordinates of each crossing. — Drop a vertical line from each intersection to the $x$-axis and read carefully.
3. $x \approx -1$ and $x \approx 3$ — [A1 for each value] In this case the solutions happen to be exact integers, but always read from the graph as instructed — do not assume integer answers.

Answer: $x \approx -1$ and $x \approx 3$

5. Guided Practice

Now that you have seen the method in action, try these problems. For each one, decide whether you need to draw a horizontal line, a slanted line, or read directly from two given curves. The rearrangement step is the one that earns the method mark, so show it clearly.

6. Exam Technique and Common Mistakes

Graphical solution questions follow a predictable pattern on GCSE Higher exam papers, and the marks are very collectable once you know the traps. The rearrangement step is almost always worth a method mark, so even if you read the graph slightly inaccurately, showing correct algebra protects your score.

7. Exam-Style Question

This final question brings together rearrangement, graph reading, and answer communication in the style you will see on an Higher-tier paper. Work through it before checking the model answer.