1MA1 · A11–A12AlgebraFoundation & HigherGrades 4–7

Quadratic Graphs

Q: Why is a quadratic graph symmetric?

Values of x equally far either side of the turning point give the same y, because of the squared term.

Q: How do I find the turning point without completing the square?

Find the line of symmetry halfway between the roots, then substitute that x value into the equation.

Q: What if the graph doesn’t cross the x-axis?

The equation ax^2 + bx + c = 0 has no real solutions.

The graph of any quadratic $y = a x^{2} + b x + c$ is a smooth U-shaped curve called a $parabola$ . Three features describe it: where it crosses the axes, and where it turns. Each feature connects to algebra you already know — roots from factorising, the turning point from completing the square — so the graph and the equation are two views of the same thing.

55 min Quadratics

The big picture

The roots of $a x^{2} + b x + c = 0$ are exactly the $x$ -intercepts of $y = a x^{2} + b x + c$ . That is why a quadratic equation can have two, one or no solutions: the parabola can cross the $x$ -axis twice, touch it once, or miss it entirely.

Edexcel tests quadratic graphs in three ways: completing a table and plotting, sketching from algebra with key points labelled, and using a graph to solve equations such as $x^{2} - 2 x - 8 = 3$ by drawing a horizontal line.

What you'll learn

Plot quadratic graphs from a table of values
Identify roots, the $y$ -intercept, the line of symmetry and the turning point
Sketch quadratic graphs from factorised and completed-square forms
Solve quadratic equations graphically and interpret the number of solutions

Plotting from a table

Substitute each $x$ value into the equation carefully, especially negative values: for $y = x^{2} - 2 x - 8$ at $x = - 3$ , $y = 9 + 6 - 8 = 7$ .

Plot the points and join them with a $smooth curve$ — never straight line segments, and never a flat bottom between two points.

The $y$ -values are symmetric about the turning point, which is a quick check on your table.

x = - 3, - 2, - 1, 0, 1, 2, 3, 4, 5

y = 7, 0, - 5, - 8, - 9, - 8, - 5, 0, 7

3The values mirror around

x = 1

AnswerRoots at

x = - 2

and

x = 4

; minimum at

(1, - 9)

Your turnSame method, new numbers — try it before moving on.

For $y = x^{2} - 2 x - 8$ , find $y$ when $x = - 3$ .

Your answer

Tip — If the $y$ -values in your table are not symmetric, recheck the substitutions before plotting — one has almost certainly gone wrong.

Key features

$Shape$ : if $a > 0$ the parabola is U-shaped with a minimum; if $a < 0$ it is ∩-shaped with a maximum.

$y-intercept$ : substitute $x = 0$ , giving $(0, c)$ .

$Roots$ : solve $a x^{2} + b x + c = 0$ . Factorised form $y = (x - r) (x - s)$ shows them directly as $x = r$ and $x = s$ .

$Line of symmetry$ and $turning point$ : halfway between the roots, or from completed-square form $y = (x + p)^{2} + q$ , which has turning point $(- p, q)$ .

x = \frac{r + s}{2}

Line of symmetry, halfway between the roots.

Your turnSame method, new numbers — try it before moving on.

Find the turning point of $y = x^{2} - 4 x + 1$ .

Your answer

Type coordinates like (2, -3).

Why this works

For $y = x^{2} - 2 x - 8 = (x - 4) (x + 2)$ , the roots $- 2$ and $4$ average to $1$ , so the line of symmetry is $x = 1$ and the turning point is at $y = 1 - 2 - 8 = - 9$ . No table needed.

Sketching

A sketch shows the correct shape and labels the key points; it does not need to be to scale.

Label every point where the curve meets the axes, and the turning point, with coordinates.

a < 0

, so the curve is ∩-shaped.

y

-intercept

(0, - 5)

3Roots:

- (x^{2} - 6 x + 5) = - (x - 1) (x - 5)

, so

x = 1

and

x = 5

4Symmetry at

x = 3

; maximum

y = - 9 + 18 - 5 = 4

Answer∩-shaped, crossing

(1, 0)

and

(5, 0)

, through

(0, - 5)

, maximum at

(3, 4)

Your turnSame method, new numbers — try it before moving on.

Where does $y = x^{2} - 7 x + 10$ cross the $x$ -axis? Give the $x$ -values.

Your answer

Separate with a comma.

Solving equations graphically

The solutions of $a x^{2} + b x + c = 0$ are the $x$ -coordinates where the curve crosses the $x$ -axis.

To solve $a x^{2} + b x + c = k$ , draw the horizontal line $y = k$ and read the $x$ -coordinates of the intersections.

To solve a rearranged equation, first rewrite it so one side matches the plotted curve.

1Draw

y = 4

2It meets the curve at about

x = - 2.6

and

x = 4.6

Answer

x \approx - 2.6

x \approx 4.6

(exactly

1 \pm 13

, i.e.

- 2.61

and

4.61

)

Tip — Graphical solutions are estimates. Give them to the accuracy the graph allows, usually 1 decimal place.

Think like an examiner

Join plotted points with a smooth curve through every point.
Label axis intercepts and the turning point with coordinates on sketches.
Show the line $y = k$ used to solve an equation graphically.

Parabolas

y = (x - r) (x - s)

Roots at

x = r

and

x = s

y = (x + p)^{2} + q

Turning point

(- p, q)

(0, c)

y-intercept of

a x^{2} + b x + c

Watch out for these

Evaluating

(- 3)^{2} - 2 (- 3)

- 9 + 6

(- 3)^{2} = 9

and

- 2 (- 3) = + 6

Joining points with straight lines or a flat bottom.

Draw one smooth curve.

Reading the

y

-intercept as a root.

Roots are where

y = 0

, on the

x

-axis.

Sketching

y = - x^{2} + \dots

as U-shaped.

A negative

x^{2}

coefficient gives a ∩ shape.

Stretch yourself

The graph of $y = x^{2} + 4 x + k$ touches the $x$ -axis at exactly one point. Find $k$ , and the coordinates of that point.

Hint — Touching the axis once means the equation $x^{2} + 4 x + k = 0$ has one repeated root. Completing the square helps.

Questions students ask

Why is a quadratic graph symmetric?

Values of

x

equally far either side of the turning point give the same

y

, because of the squared term.

How do I find the turning point without completing the square?

Find the line of symmetry halfway between the roots, then substitute that

x

value into the equation.

What if the graph doesn’t cross the

x

-axis?

The equation

a x^{2} + b x + c = 0

has no real solutions.

Key takeaways

$a > 0$ gives a U-shape; $a < 0$ gives a ∩-shape.
The $y$ -intercept is $(0, c)$ ; roots are the solutions of $y = 0$ .
The line of symmetry is halfway between the roots and passes through the turning point.
Completed-square form gives the turning point directly.
Solve $f (x) = k$ graphically by drawing $y = k$ .

Exam-style questions

Practise it the way it is examined

3 original questions · 6 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 43 marks

This question is about $y = x^{2} + 2 x - 15$ .

(a)Find the roots.[2]
(b)Write down the coordinates of the $y$ -intercept.[1]

Question 2Grade 51 mark

A quadratic graph crosses the $x$ -axis at $- 3$ and $7$ . Write down the equation of its line of symmetry.[1]

Question 3Grade 72 marks

The graph of $y = x^{2} - 4 x$ has been drawn. What straight line should be drawn to solve $x^{2} - 5 x + 2 = 0$ ?[2]

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Quadratic Graphs

Plotting from a table

Key features

Sketching

Solving equations graphically

Practise it the way it is examined

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Quadratic Graphs

Plotting from a table

Key features

Sketching

Solving equations graphically

Practise it the way it is examined

Skill drill

Practise this topic

Related topics