Pure 2.3 - Quadratic functions and graphs | Edexcel A-Level Mathematics

A quadratic's graph, roots and minimum or maximum are different views of the same expression. You will complete the square, interpret the discriminant, solve quadratic equations and recognise quadratics hidden inside powers, exponentials, trigonometric functions and logarithms.

Quadratic graph features

Can $2x^2-8x+13$ ever be negative? At $x=0$ it is $13$ , and at $x=2$ it is $5$ , but checking a few values cannot settle what happens for every real input. We need the shape and the algebra to agree.

You already know

A square is non-negative: $t^2\ge0$ for every real $t$ . A root of an equation is an input making it true; factorisation can turn an equation into a zero product.

The notation $f(x)$ means the output at input $x$ ; it does not mean $f$ multiplied by $x$ . The function $f(x)=ax^2+bx+c$ , with $a\ne0$ , is quadratic, and its graph is a parabola. The coefficient $a$ decides whether it opens upwards, with a minimum, or downwards, with a maximum.

Root and intercept

A real root of $f(x)=0$ gives an $x$ -intercept $(x, 0)$ . The $y$ -intercept is $(0, c)$ , since $f(0)=c$ . A turning point is where the parabola changes from decreasing to increasing, or conversely.

The parabola can cross the horizontal axis twice, touch it once, or miss it entirely. A touching point represents a repeated root, counted twice algebraically but appearing at only one coordinate. A quadratic needs $a\ne0$ : $0x^2+3x+2$ is linear.

Key idea

Roots locate horizontal-axis intersections; the turning point locates the minimum or maximum. They answer different questions.

Use the coefficient and constant to read the following features without solving an equation.

Completing the square

Start with $x^2-4x$ . Since $(x-2)^2=x^2-4x+4$ , subtracting $4$ restores the expression: $x^2-4x=(x-2)^2-4$ . Completing the square makes a quadratic's symmetry visible.

Completed-square form

For $a\ne0$ ,

ax^2+bx+c=a\left(x+\frac b{2a}\right)^2+c-\frac{b^2}{4a}

In the form $a(x-p)^2+q$ , the turning point is $(p, q)$ and the axis of symmetry is $x=p$ .
Not in the formula booklet: learn it.

When $a\ne1$ , factor it from the quadratic and linear terms before halving the inner coefficient. Any correction made inside that bracket is multiplied by $a$ .

Worked example 1: a non-unit leading coefficient

Question. Complete the square for $f(x)=2x^2-8x+3$ , and state its turning point.

Plan. Factor out $2$ , complete the inner square, then collect constants.

\begin{aligned} f(x)&=2(x^2-4x)+3 &&\text{factor out }2\\ &=2\big((x-2)^2-4\big)+3 &&\text{restore the added }4\\ &=2(x-2)^2-5 &&\text{collect constants} \end{aligned}

Why this step? Subtracting $4$ inside the bracket subtracts $8$ from the whole expression.

Answer. $2(x-2)^2-5$ , with minimum turning point $(2, -5)$ .

Check. Expansion gives $2x^2-8x+8-5=2x^2-8x+3$ .

[DIAGRAM: asset_slug: p1_2_3_work_with_quadratic_functions_and_their_graphs__diagram_02; asset_name: Completed square and turning point; caption: Completing the square exposes the axis and minimum; description: The graph y equals 2 times x minus 2 squared minus 5 has turning point (2, minus 5), symmetry axis x equals 2 and y-intercept (0, 3). Its roots are approximately 0.419 and 3.58.]

Change only the constant from $3$ to $13$ . The graph rises by $10$ , leaving the symmetry axis unchanged.

Worked example 2: a vertical shift

Question. Complete the square for $2x^2-8x+13$ , and find its minimum.

Plan. Reuse the same inner square and change the final constant.

\begin{aligned} 2x^2-8x+13&=2\big((x-2)^2-4\big)+13 &&\text{complete the square}\\ &=2(x-2)^2+5\ge5 &&\text{a square is non-negative} \end{aligned}

Why this step? The bound is attained when $x=2$ , so it is the minimum, not merely a lower estimate.

Answer. Minimum $5$ , attained at $x=2$ .

Check. $2(2)^2-8(2)+13=5$ ; values at $x=1$ and $x=3$ are both $7$ .

The sign inside the square is easy to misread.

Watch out

Wrong: the turning point of $(x+5)^2-4$ is $(5, -4)$ . The square is zero at $x=-5$ , so the turning point is $(-5, -4)$ .

Apply the same square-and-correct method with a new linear coefficient.

The discriminant

The quadratic formula contains $\sqrt{b^2-4ac}$ . A positive number under that root gives two real choices, zero makes them coincide, and a negative number gives no real value. This lets you count roots without solving for them.

Discriminant

For $ax^2+bx+c=0$ , $a\ne0$ , define

D=b^2-4ac

$D>0$ : two distinct real roots; $D=0$ : a repeated real root; $D<0$ : no real roots.
Not in the formula booklet: learn it.

These are also statements about the graph's intersections with the horizontal axis. A repeated root still solves the equation; it is not “no roots”.

[DIAGRAM: asset_slug: p1_2_3_work_with_quadratic_functions_and_their_graphs__diagram_01; asset_name: Three discriminant cases; caption: The discriminant counts real roots; description: Three upward parabolas show two crossings for D positive, one touching point for D zero and no crossing for D negative.]

Worked example 3: no real roots

Question. Determine the number of real roots of $x^2-6x+10=0$ .

Plan. Calculate the discriminant and interpret its sign.

\begin{aligned} D&=(-6)^2-4(1)(10) &&\text{include the sign of }b\\ &=36-40=-4<0 &&\text{evaluate} \end{aligned}

Why this step? Counting roots needs only the sign, not the formula's two values.

Answer. There are no real roots.

Check. $x^2-6x+10=(x-3)^2+1\ge1$ , so it cannot equal zero.

Change only the constant from $10$ to $9$ . The graph now just reaches the axis.

Worked example 4: a repeated root

Question. Determine the number of distinct real roots of $x^2-6x+9=0$ .

Plan. Recalculate the discriminant.

\begin{aligned} D&=(-6)^2-4(1)(9) &&\text{substitute}\\ &=36-36=0 &&\text{repeated-root condition} \end{aligned}

Why this step? The two signs in the quadratic formula give the same value when its square root is zero.

Answer. One distinct real root, repeated.

Check. $x^2-6x+9=(x-3)^2$ , so the root is $3$ twice.

A negative discriminant by itself does not tell you whether the curve stays above or below the axis; the sign of $a$ supplies that information.

Watch out

Wrong: “ $D=0$ , so there are two distinct roots.” Right: the roots coincide. Use $D>0$ for two distinct real roots.

A parameter question reverses the process: set the required discriminant condition first.

Solving quadratic equations

A factorisation such as $(x-2)(3x+4)=0$ gives two linear equations, because a zero product has at least one zero factor. If the brackets are not obvious, the quadratic formula works for every quadratic with real roots.

Quadratic formula

For $ax^2+bx+c=0$ , with $a\ne0$ and $b^2-4ac\ge0$ ,

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Not in the formula booklet: learn it.

Put the equation equal to zero before reading $a$ , $b$ , $c$ . Factorisation is often shortest; completing the square is equally valid and reveals the turning point at the same time.

Worked example 5: factorisation

Question. Solve $3x^2-2x-8=0$ .

Plan. Factorise, then set each factor equal to zero.

\begin{aligned} 3x^2-2x-8&=(3x+4)(x-2) &&\text{cross-products give }-2x\\ 3x+4&=0\quad\text{or}\quad x-2=0 &&\text{zero product} \end{aligned}

Why this step? Dividing by one factor could discard the root that makes it zero.

Answer. $x=-\frac43$ or $x=2$ .

Check. Substitution gives $16/3+8/3-8=0$ and $12-4-8=0$ .

Change only the constant from $-8$ to $-9$ . The roots are now surds, so the formula is convenient.

Worked example 6: exact surd roots

Question. Solve $3x^2-2x-9=0$ exactly.

Plan. Substitute $a=3$ , $b=-2$ , $c=-9$ in the formula and simplify.

\begin{aligned} x&=\frac{2\pm\sqrt{(-2)^2-4(3)(-9)}}6 &&\text{quadratic formula}\\ &=\frac{2\pm\sqrt{112}}6=\frac{1\pm2\sqrt7}{3} &&\text{extract a square factor} \end{aligned}

Why this step? Keeping the root exact avoids losing information through rounding.

Answer. $x=(1+2\sqrt7)/3$ or $x=(1-2\sqrt7)/3$ .

Check. Rearranging gives $(3x-1)^2=28$ , satisfied by both answers.

For comparison, completing the square in $2x^2-4x-1=0$ gives $2(x-1)^2=3$ , hence $x=1\pm\sqrt6/2$ . A calculator's polynomial solver would return approximately $2.22$ and $-0.225$ . Enter coefficients $2$ , $-4$ , $-1$ , retaining the negative signs; substitute outputs back to check. Calculator roots are acceptable when permitted, but do not replace required algebraic working or exact forms.

Exam tip

If the question excludes solutions relying entirely on calculator technology, display factorisation, square completion or formula substitution. A calculator is still useful for checking your algebra.

Use an efficient exact method for this new quadratic.

Quadratics in a function of the unknown

The expression $x^4-5x^2+4$ contains the square of $x^2$ . Setting $u=x^2$ reveals an ordinary quadratic. Solve for $u$ , then solve for $x$ ; stopping at $u$ leaves the original question unanswered.

You already know

$4^x=2^{2x}=(2^x)^2$ . A logarithm reverses a power: $2^x=3$ means $x=\log_2 3$ . Since $2^x>0$ , it can never equal a negative number.

Substitution preserves the repeated expression. Check its range: if $u=x^2$ , then $u\ge0$ ; if $u=2^x$ , then $u>0$ . Neither statement is a formula supplied in the booklet.

Worked example 7: a quadratic in a square

Question. Solve $x^4-5x^2+4=0$ .

Plan. Put $u=x^2$ , solve, then take both square roots of each positive value.

\begin{aligned} u^2-5u+4&=(u-1)(u-4)=0 &&\text{substitute and factorise}\\ x^2&=1\quad\text{or}\quad x^2=4 &&\text{back-substitute} \end{aligned}

Why this step? Each positive value of $x^2$ gives both a positive and a negative $x$ .

Answer. $x=\pm1$ , $\pm2$ .

Check. The original gives $1-5+4=0$ when $x^2=1$ , and $16-20+4=0$ when $x^2=4$ .

Change only the constant to $-6$ . One temporary root will now fail its range condition.

Worked example 8: rejecting an impossible square

Question. Solve $x^4-5x^2-6=0$ over the reals.

Plan. Use $u=x^2\ge0$ , then reject negative values of $u$ .

\begin{aligned} u^2-5u-6&=(u-6)(u+1)=0 &&\text{quadratic in }u\\ x^2&=6\quad\text{or}\quad x^2=-1 &&\text{back-substitute} \end{aligned}

Why this step? A real square cannot be $-1$ , although $-1$ solves the temporary quadratic.

Answer. $x=\pm\sqrt6$ .

Check. $36-30-6=0$ for either sign.

The same substitution idea works when the repeated expression is an exponential. Here is a second pair so you can see the different back-substitution step.

Worked example 9: an exponential quadratic

Question. Solve $4^x-7(2^x)+12=0$ exactly.

Plan. Put $u=2^x>0$ , factorise, then reverse the exponential.

\begin{aligned} u^2-7u+12&=(u-3)(u-4)=0 &&\text{since }4^x=u^2\\ 2^x&=3\quad\text{or}\quad 2^x=4 &&\text{back-substitute} \end{aligned}

Why this step? The temporary roots are outputs of $2^x$ , not values of $x$ .

Answer. $x=\log_2 3$ or $x=2$ .

Check. With outputs $3$ and $4$ , the original is respectively $9-21+12=0$ and $16-28+12=0$ .

Change only the constant to $-8$ . Exponential outputs must still be positive.

Worked example 10: an impossible exponential output

Question. Solve $4^x-7(2^x)-8=0$ .

Plan. Use $u=2^x>0$ and check the quadratic roots against that condition.

\begin{aligned} u^2-7u-8&=(u-8)(u+1)=0 &&\text{factorise}\\ 2^x&=8 &&\text{reject }u=-1 \end{aligned}

Why this step? No real power of a positive base is negative.

Answer. $x=3$ .

Check. $4^3-7(2^3)-8=64-56-8=0$ .

The natural exponential $e^x$ works identically; its inverse is $\ln$ , so $e^x=k$ gives $x=\ln k$ for $k>0$ .

Watch out

Wrong: “ $u=8$ , so $x=8$ .” If $u=2^x$ , solve $2^x=8$ , giving $x=3$ . Always undo your substitution.

Try the same quadratic structure with base $e$ .

Quadratics in trigonometric and logarithmic functions

A quadratic can hide in $\sin x$ or $\ln x$ just as it hides in $x^2$ . The quadratic algebra does not change; the last step does. First make the small amount of function knowledge explicit.

You already know

In degrees, $\sin30^\circ=\sin150^\circ=1/2$ , $\sin90^\circ=1$ , and $-1\le\sin x\le1$ . On one full turn, sine is positive in the first two quadrants and negative in the last two. For $x>0$ , $\ln x=t$ means $x=e^t$ ; a logarithm's output may be negative.

For $\sin x=s$ with $|s|\le1$ , use the reference angle $\alpha=\sin^{-1}|s|$ in degree mode and the quadrants to find every solution in the stated interval. For example, $s=-1/2$ gives $180^\circ+30^\circ$ and $360^\circ-30^\circ$ . The inverse key alone does not list both. For logs, check the argument is positive, not the output. These inverse definitions and sine values are not supplied in the formula booklet.

Worked example 11: a quadratic in sine

Question. Solve $2\sin^2x-3\sin x+1=0$ for $0^\circ\le x<360^\circ$ .

Plan. Put $u=\sin x$ , solve the quadratic, then find every angle in the interval.

\begin{aligned} 2u^2-3u+1&=(2u-1)(u-1)=0 &&\text{quadratic in sine}\\ \sin x&=\frac12\quad\text{or}\quad\sin x=1 &&\text{back-substitute} \end{aligned}

Why this step? A sine value normally corresponds to two angles on a full turn; the maximum value occurs once.

Answer. $x=30^\circ$ , $90^\circ$ , $150^\circ$ .

Check. Sine outputs $1/2$ and $1$ give $1/2-3/2+1=0$ and $2-3+1=0$ .

Change only the constant from $1$ to $-2$ . One quadratic root is now outside sine's range.

Worked example 12: checking the sine range

Question. Solve $2\sin^2x-3\sin x-2=0$ for $0^\circ\le x<360^\circ$ .

Plan. Use $u=\sin x$ and reject $|u|>1$ .

\begin{aligned} 2u^2-3u-2&=(2u+1)(u-2)=0 &&\text{factorise}\\ \sin x&=-\frac12 &&\text{reject }u=2\\ x&=180^\circ+30^\circ\quad\text{or}\quad360^\circ-30^\circ &&\text{negative sine} \end{aligned}

Why this step? $u=2$ is an algebraic root but cannot be a real sine output.

Answer. $x=210^\circ$ , $330^\circ$ .

Check. Each gives sine $-1/2$ , so the original is $1/2+3/2-2=0$ .

Now use $u=\ln x$ . This time $u$ has no upper or lower bound, but $x$ must be positive. Logs reverse exponentials, so there is one positive input for each real output.

Worked example 13: a quadratic in a logarithm

Question. Solve $(\ln x)^2-3\ln x+2=0$ .

Plan. State $x>0$ , put $u=\ln x$ , then exponentiate each root.

\begin{aligned} u^2-3u+2&=(u-1)(u-2)=0 &&\text{substitute}\\ \ln x&=1\quad\text{or}\quad\ln x=2 &&\text{back-substitute}\\ x&=e\quad\text{or}\quad x=e^2 &&\text{inverse functions} \end{aligned}

Why this step? $\ln x=2$ asks which input has logarithm $2$ ; it does not give $x=2$ .

Answer. $e$ and $e^2$ , both in the domain $x>0$ .

Check. Their logarithms are $1$ and $2$ , giving $1-3+2=0$ and $4-6+2=0$ .

Change only the logarithm's argument from $x$ to $x-1$ . The domain and final inverse step must change together.

Worked example 14: checking a shifted logarithm's domain

Question. Solve $(\ln(x-1))^2-3\ln(x-1)+2=0$ .

Plan. Require $x>1$ , then use $u=\ln(x-1)$ .

\begin{aligned} (u-1)(u-2)&=0 &&\text{same quadratic}\\ x-1&=e\quad\text{or}\quad x-1=e^2 &&\text{undo the logarithm}\\ x&=1+e\quad\text{or}\quad x=1+e^2 &&\text{restore the shift} \end{aligned}

Why this step? Exponentiating finds the whole argument $x-1$ , not $x$ .

Answer. $1+e$ , $1+e^2$ , both greater than $1$ .

Check. Subtracting $1$ gives positive arguments whose logs are $1$ and $2$ .

The same algebra can yield a negative logarithm, which is perfectly valid: $\ln x=-1$ gives $x=e^{-1}>0$ .

Watch out

Reject $\sin x=2$ , because the output is outside $[-1, 1]$ . Do not reject $\ln x=-1$ : only the log's argument must be positive.

This final check samples both back-substitution rules with new quadratic coefficients.

Choosing the method

Choose your representation to match your target. Factor form exposes roots, completed-square form exposes bounds, and the discriminant counts real roots. This is more reliable than trying the same tool on every question.

Target	Useful first move
Roots	Factorise, use the formula, complete the square or use a permitted calculator solver
Turning point or bound	Complete the square
Number of real roots	Calculate the discriminant
Repeated expression	Substitute, solve, back-substitute and check conditions

A surprising fact is that a positive value at every tested integer says little about all real inputs: $x^2-x+1/5$ is positive at every integer, yet at $x=1/2$ it is $-1/20$ . Indeed, an integer $n$ satisfies $n\le0$ or $n\ge1$ , so $n(n-1)\ge0$ and $n^2-n+1/5>0$ . Between-integer behaviour needs an argument.

Worked example 15: proving positivity

Question. Show that $2x^2-8x+13>0$ for every real $x$ .

Plan. Complete the square and give a strict positive lower bound.

\begin{aligned} 2x^2-8x+13&=2(x-2)^2+5 &&\text{complete the square}\\ &\ge5>0 &&\text{since }(x-2)^2\ge0 \end{aligned}

Why this step? The added $5$ , rather than the square alone, makes the result strictly positive.

Answer. The expression is positive for all real $x$ .

Check. Its smallest value is attained at $x=2$ and is $5$ , answering the opening question.

Change only the constant from $13$ to $8$ . The distinction between positive and non-negative now matters.

Worked example 16: a boundary case

Question. Determine whether $2x^2-8x+8$ is positive for every real $x$ .

Plan. Complete the square and inspect equality.

\begin{aligned} 2x^2-8x+8&=2(x-2)^2 &&\text{complete the square}\\ &\ge0 &&\text{equality at }x=2 \end{aligned}

Why this step? A square can be zero, so non-negative does not imply strictly positive.

Answer. No: it is non-negative for every real input but equals zero at $x=2$ .

Check. $2(2)^2-8(2)+8=0$ .

A discriminant argument can also prove positivity when you include $a>0$ : an upward parabola with no real roots stays above the axis.

Exam tip

A “show that” proof needs a general reason and a conclusion. Testing several inputs cannot establish a claim for every real number.

Use the same completed-square reasoning to establish a new printed result.

Quick Recap

$a(x-p)^2+q$ has turning point $(p, q)$ ; inspect the sign of $a$ .
$D=b^2-4ac$ distinguishes two real roots, a repeated root and no real roots.
Choose a solving method to fit the required answer and show required algebra.
A substituted quadratic is unfinished until you return to the original variable and check its domain, range and interval.

The opening expression can never be negative: $2x^2-8x+13=2(x-2)^2+5\ge5$ .

Last Accessed

Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 2.3 - Quadratic functions and graphs

Quadratic graph features

Completing the square

The discriminant

Solving quadratic equations

Quadratics in a function of the unknown

Quadratics in trigonometric and logarithmic functions

Choosing the method

Quick Recap

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