Maths · 3 Functions

3.4 Finding a quadratic function using given information

Before you start

Expanding brackets, including ${(x - 3)}^{2}$ and $(x - 2) (x + 5)$ .
That a quadratic graph is a parabola with one vertex (3.1).
Solving a simple linear equation for one unknown (2.5).

Syllabus objectives covered (1)

Find a quadratic function given: (a) vertex and another point (b) x-intercepts and a point (c) vertex or x-intercepts in the case where a = 1.

Part 1 of 3

Two forms, and choosing the right one

Everything in this subtopic rests on writing a quadratic in whichever form matches the information you are given. Pick the wrong form and the algebra becomes far harder than it needs to be.
Vertex form, when you are told the vertex:
$y = a {(x - h)}^{2} + k$
This has vertex $(h, k)$ — and note the minus sign in the bracket. A vertex at $(3, - 4)$ gives $y = a {(x - 3)}^{2} - 4$ ; a vertex at $(- 3, - 4)$ gives $y = a {(x + 3)}^{2} - 4$ .
That sign flip is the single most common error in this subtopic. The number in the bracket always has the opposite sign to the $x$ -coordinate of the vertex.
Factorised form, when you are told where the curve crosses the $x$ -axis:
$y = a (x - p) (x - q)$
This has $x$ -intercepts at $x = p$ and $x = q$ , with the same sign-flip rule. Intercepts at $2$ and $- 5$ give $y = a (x - 2) (x + 5)$ .
So read the question first and ask which you have been given: a vertex, or the intercepts. That decision is most of the work.

Part 2 of 3

The method

Whichever form you use, the procedure is the same three steps.
Step 1. Write down the form with $h$ and $k$ , or $p$ and $q$ , filled in from the given points. Leave $a$ as a letter.
Step 2. Substitute the coordinates of the extra point — the one that is neither the vertex nor an intercept — and solve for $a$ .
Step 3. Write the full function, expanding into $y = a x^{2} + b x + c$ if the question asks for that form.
The extra point is what makes the answer unique. Infinitely many parabolas share a given vertex; only one also passes through a specified extra point.
**The special case $a = 1$ .** If the question says the coefficient of $x^{2}$ is 1, no extra point is needed — you can write the answer straight down from the vertex or the intercepts.
With intercepts $2$ and $- 5$ and $a = 1$ : $y = (x - 2) (x + 5)$ , which expands to $y = x^{2} + 3 x - 10$ .
Always check your answer by substituting the given points back in. A sign error in a bracket shows up immediately.

Part 3 of 3

Reading a graph for the information

Often the information comes from a diagram rather than words, so you have to spot which form applies.
If the diagram marks the turning point, use vertex form. If it marks two crossings on the $x$ -axis, use factorised form.
The $y$ -intercept is frequently the 'extra point'. It is the point $(0, c)$ , and substituting $x = 0$ is the easiest substitution there is.
The sign of $a$ is a free check. If the parabola opens upwards, $a$ must come out positive; if downwards, negative. If your $a$ has the wrong sign, you have made an error.
A parabola that touches the $x$ -axis at exactly one point has a repeated root there, so $p = q$ and the form becomes $y = a {(x - p)}^{2}$ — which is also its vertex form, with the vertex on the axis.

Worked examples

Example 1. A quadratic curve has vertex $(3, - 4)$ and passes through the point $(5, 8)$ . Find the function in the form $y = a x^{2} + b x + c$ .

Show the working

The vertex is given, so use vertex form $y = a {(x - h)}^{2} + k$ with $h = 3$ and $k = - 4$ .
$y = a {(x - 3)}^{2} - 4$
Substitute the extra point $(5, 8)$ to find $a$ .
$8 = a {(5 - 3)}^{2} - 4$
Simplify and solve.
$8 = 4 a - 4 \Rightarrow 4 a = 12 \Rightarrow a = 3$
So the function is $y = 3 {(x - 3)}^{2} - 4$ . Now expand as the question asks.
$y = 3 (x^{2} - 6 x + 9) - 4 = 3 x^{2} - 18 x + 23$
Check against the vertex: at $x = 3$ , $y = 27 - 54 + 23 = - 4$ . Correct. And $a = 3 > 0$ , so the parabola opens upwards, consistent with $(3, - 4)$ being a minimum.

Answer: $y = 3 x^{2} - 18 x + 23$ .

Example 2. A parabola crosses the $x$ -axis at $x = - 1$ and $x = 4$ , and passes through $(0, 6)$ . Find its equation.

Show the working

The $x$ -intercepts are given, so use factorised form with the signs flipped.
$y = a (x + 1) (x - 4)$
Substitute the extra point $(0, 6)$ .
$6 = a (0 + 1) (0 - 4) = - 4 a$
Solve for $a$ .
$a = - \frac{6}{4} = - 1.5$
Write the function.
$y = - 1.5 (x + 1) (x - 4)$
Sense-check the sign: $a$ is negative, so the parabola opens downwards. With a positive $y$ -intercept between two roots, that is exactly what it should do.

Answer: $y = - 1.5 (x + 1) (x - 4)$ , or expanded, $y = - 1.5 x^{2} + 4.5 x + 6$ .

Example 3. A quadratic function has $a = 1$ and vertex $(- 2, 5)$ . Find the function in the form $y = x^{2} + b x + c$ .

Show the working

Because $a = 1$ is given, no extra point is needed — the vertex alone determines the curve.
Use vertex form, remembering the sign flip: a vertex $x$ -coordinate of $- 2$ gives $(x + 2)$ in the bracket.
$y = {(x + 2)}^{2} + 5$
Expand the bracket.
$y = x^{2} + 4 x + 4 + 5$
Simplify.
$y = x^{2} + 4 x + 9$
Check: at $x = - 2$ , $y = 4 - 8 + 9 = 5$ . The vertex is where it should be.

Answer: $y = x^{2} + 4 x + 9$ .

Where people go wrong

Mistake: Writing $y = a {(x + h)}^{2} + k$ for a vertex at $(h, k)$ .

Instead: The form is $y = a {(x - h)}^{2} + k$ , so the bracket contains the opposite sign to the vertex's $x$ -coordinate.

A vertex at $(3, - 4)$ needs $(x - 3)$ ; a vertex at $(- 3, - 4)$ needs $(x + 3)$ . Getting this backwards moves the curve to the wrong place entirely.

Mistake: Using $y = a (x + p) (x + q)$ for intercepts at $p$ and $q$ .

Instead: It is $y = a (x - p) (x - q)$ .

Setting each bracket to zero must give back the intercepts. With $(x - 2)$ the root is $x = 2$ , which is the point of the form.

Mistake: Forgetting the extra point and leaving $a$ unknown.

Instead: Substitute the third given point to find $a$ , unless the question states $a = 1$ .

The vertex or the intercepts alone describe a whole family of parabolas. The extra point picks out one.

Mistake: Expanding ${(x - 3)}^{2}$ as $x^{2} - 9$ or $x^{2} + 9$ .

Instead: ${(x - 3)}^{2} = x^{2} - 6 x + 9$ .

Squaring a bracket means multiplying it by itself, which produces a middle term. Write it out as $(x - 3) (x - 3)$ if you are unsure.

Mistake: Multiplying out but forgetting that $a$ multiplies every term.

Instead: $3 (x^{2} - 6 x + 9) = 3 x^{2} - 18 x + 27$ — all three terms.

A partly-applied multiplier gives a curve through the right vertex but the wrong shape.

Check yourself

Write down the vertex of $y = 2 {(x - 5)}^{2} + 3$ .
Answer
$(5, 3)$ .
Write down the vertex of $y = - {(x + 1)}^{2} - 7$ .
Answer
$(- 1, - 7)$ .
A parabola with $a = 1$ crosses the $x$ -axis at $3$ and $- 2$ . Write down its equation in factorised form and expand it.
Answer
$y = (x - 3) (x + 2) = x^{2} - x - 6$ .
A curve has vertex $(1, 2)$ and passes through $(3, 10)$ . Find $a$ .
Answer
$10 = a {(3 - 1)}^{2} + 2$ , so $4 a = 8$ and $a = 2$ .
Why is an extra point needed as well as the vertex?
Answer
Infinitely many parabolas share the same vertex, differing in how wide or narrow they are. The extra point fixes $a$ and so picks out one.
A parabola touches the $x$ -axis at $x = 4$ only. What does this tell you about its factorised form?
Answer
It has a repeated root, so $y = a {(x - 4)}^{2}$ , and its vertex is at $(4, 0)$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. The diagram shows a parabola with vertex $(- 1, - 8)$ passing through the point $(1, 0)$ .

(a) Write down the equation of the curve in the form $y = a {(x - h)}^{2} + k$ , leaving $a$ as an unknown. 2 marks No calculator

Mark scheme

B1 (x + 1)² seen
B1 y = a(x + 1)² - 8

(b) Find the value of $a$ . 2 marks No calculator

Mark scheme

M1 0 = a(1 + 1)² - 8 oe
A1 a = 2

(c) Show that the equation can be written as $y = 2 x^{2} + 4 x - 6$ . 2 marks No calculator

Mark scheme

M1 2(x² + 2x + 1) - 8 seen; FT their a
A1 reaches the given result with no errors; nfww

Question 2. A quadratic curve crosses the $x$ -axis at $x = - 3$ and $x = 5$ , and crosses the $y$ -axis at $(0, - 30)$ .

(a) Find the equation of the curve, giving your answer in factorised form. 3 marks No calculator

Mark scheme

M1 y = a(x + 3)(x - 5) seen
M1 -30 = a(3)(-5) oe
A1 y = 2(x + 3)(x - 5) oe

(b) Write down whether the vertex of this curve is a maximum or a minimum, and give a reason. 1 mark No calculator

Mark scheme

B1 minimum, because a is positive so the parabola opens upwards; FT their a

Question 3. A quadratic function has coefficient of $x^{2}$ equal to $1$ and a vertex at $(4, - 9)$ .

(a) Find the function in the form $y = x^{2} + b x + c$ . 3 marks No calculator

Mark scheme

M1 y = (x - 4)² - 9 seen
M1 expands (x - 4)² to x² - 8x + 16
A1 y = x² - 8x + 7

(b) Solve $y = 0$ for this function. 2 marks No calculator

Mark scheme

M1 (x - 1)(x - 7) seen, or correct use of the quadratic formula; FT their (a)
A1 x = 1 and x = 7

Words to know

Vertex form: $y = a {(x - h)}^{2} + k$ , whose vertex is at $(h, k)$ .
Factorised form: $y = a (x - p) (x - q)$ , whose $x$ -intercepts are $p$ and $q$ .
Vertex: The turning point of a parabola — its minimum if $a > 0$ , its maximum if $a < 0$ .
$x$ -intercept: A point where the curve crosses the $x$ -axis, so $y = 0$ .
Repeated root: A root occurring twice, where the parabola touches the $x$ -axis instead of crossing it.

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