GCSE · Maths · AQA · Spec 8300 · Foundation

Approximate solutions of quadratics from a graph

You don’t always need algebra to solve a quadratic. If you have its graph, the answers are already drawn on it — you just have to read them off.

Where is x² − 2x − 4 equal to 0?

-3-1135-6061218xy(1, -5)

x: 1. y: -5

The point starts at the bottom of the U, where y = −5. Slide it right and watch y: find where it changes from negative to positive. That’s a crossing. Now find the one on the left.

Red line: y = 0: Every point on the red line has a height of exactly 0 — it is the x-axis. Wherever the curve y = x² − 2x − 4 crosses it, x² − 2x − 4 = 0. So the solutions of the equation are the x-values of those crossing points. (The x-values are numbered along the bottom edge.)

Exam line: The curve crosses y = 0 twice, so x² − 2x − 4 = 0 has two solutions: x ≈ −1.2 and x ≈ 3.2. They’re approximate because the crossings fall between the numbers — you judge them by eye.
Watch out: The bottom of the U, (1, −5), is not a solution: its height is −5, not 0. Nor is (0, −4), where the curve cuts the y-axis — there x = 0 and y = −4. Solutions are x-values where y = 0.

Your turn to choose

When the equation doesn’t equal 0

Use the graph of y = x² − 2x − 4 above to solve x² − 2x − 7 = 0.

  1. x² − 2x − 7 = 0
  2. missing step
Which line is step 2?

Predict, then check

The lowest point of y = x² − 2x − 4 is (1, −5). Picture a horizontal line at y = −7 on the graph above.

How many solutions does x² − 2x − 4 = −7 have?

Check the working

Where does this answer lose a mark?

Use the graph of y = x² − 2x − 4 to solve x² − 2x − 4 = 4.

A student’s answer — which line goes wrong?

WHAT YOU'VE LEARNED

A quick recap of today's lesson.

An equation asks a question. Draw the curve, and the answer is sitting on the picture — you just need to know where to look.

What you need to know

  • The solutions of ax² + bx + c = 0 are the x-coordinates of the points where the graph of y = ax² + bx + c meets the x-axis (the line y = 0) — whether it crosses the axis or just touches it at the turning point.
  • To solve ax² + bx + c = k, draw the horizontal line y = k and read the x-coordinates where it meets the curve.
  • If the equation doesn’t match the graph’s equation, add or subtract the same number on both sides until the left side does. The right-hand side then tells you which line to draw.
  • A horizontal line can cross the curve twice, touch it once at the turning point, or miss it, so there are two, one or no solutions.
  • Graph readings are approximate: you judge each crossing by eye, usually between two numbers on the scale.

The big picture

The graph of y = ax² + bx + c pairs every x with its y. To solve ax² + bx + c = 0, find where y = 0: the points where the curve meets the x-axis, whether it crosses it or just touches it. Their x-coordinates are the solutions. To solve ax² + bx + c = k, draw the line y = k and read the x-coordinates where it meets the curve, rearranging first if the equation doesn’t match the graph. The line can meet the curve twice, once or not at all. Readings are approximate, because you judge the crossings by eye.

Key points

1Solutions are x-values. Read down from each crossing to the x-axis — never give the y-value of the point.
2The y-intercept and the turning point are not answers in themselves. A point only gives a solution if it sits on the line you drew — and even then, the solution is its x-coordinate.
3Count the crossings before you write your answer, so you never stop at one solution when there are two.
4You can check a reading by substituting it into the equation: the left side should come out close to the right-hand side, but rarely exactly equal.

Worked example

Problem

Use the graph of y = x² − 2x − 4 to find approximate solutions of x² − 2x − 4 = 0. Then check one of your answers.

⚠ Watch out

Reading the wrong feature. For x² − 2x − 4 = 0, students give the y-intercept (−4), the bottom of the U (x = 1) or a y-value. The equation asks where y = 0, so the answers are x-values on the x-axis — and here there are two of them.

🧠

Memory hook

Draw the line, find where it hits, drop down to x. Line → hit → drop.

✓

Check yourself

Use the graph of y = x² − 2x − 4 to solve x² − 2x − 4 = −1. (Answer: x = −1 and x = 3.)

Flashcards

(6)
On the graph of y = ax² + bx + c, where are the solutions of ax² + bx + c = 0?
They are the x-coordinates of the points where the curve meets the x-axis — crossing it, or just touching it at the turning point — because y = 0 there.
How do you use the graph of y = ax² + bx + c to solve ax² + bx + c = k?
Draw the horizontal line y = k and read the x-coordinates of the points where it meets the curve.
The equation you need to solve doesn’t match the graph’s equation. What do you do first?
Add or subtract the same number on both sides until the left side is the graph’s equation. The right-hand side then tells you which horizontal line to draw.
How can a graph show that a quadratic equation has two, one or no solutions?
The horizontal line cuts the curve twice (two), touches it only at the turning point (one), or misses it altogether (none).
Why are solutions read from a graph only approximate?
You judge the crossing by eye on a drawn curve, and it usually falls between the numbers on the scale. Substituting the reading back gives a value close to, but rarely exactly, the right-hand side.
Why isn’t the y-intercept (0, c) a solution of ax² + bx + c = 0?
At the y-intercept x = 0 and y = c, not 0. Solutions come from where y = 0 — on the x-axis, not the y-axis.

Tap any card to flip it, or use Study as deck to go through them one at a time. In the full lesson these run as a spaced-repetition deck — you rate each card Hard, Good or Easy and the tricky ones keep coming back until they stick.

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