B3 - Quadratic graphs and discriminants | AQA A-Level Mathematics

Sketch quadratics from their intercepts and turning points, and use completed-square form to explain their maximum or minimum values. Then use the discriminant to count real roots and find parameter conditions, including values where an equation stops being quadratic.

Features of a quadratic graph

Can changing just the constant turn two crossings of the horizontal axis into no crossings at all? Compare $y=x^2-4x+3$ , $y=x^2-4x+4$ and $y=x^2-4x+5$ . Their shapes match, yet their meetings with the axis do not. We will explain exactly where the change happens.

Start with $y=x^2-4x-5$ . At $x=0$ , its value is $-5$ ; at $x=-1$ and $x=5$ , its value is zero. These easy calculations already locate three points on the curve.

Quadratic function

A quadratic function has the form $f(x)=ax^2+bx+c$ , with real constants $a,b,c$ and $a\ne0$ . Its graph is a parabola. Unless restricted, its domain is all real numbers. The expression $3x+2$ is not quadratic because its squared-term coefficient is zero.

If $a>0$ , both arms rise and the curve has a minimum. If $a<0$ , both arms fall and it has a maximum. The vertex, or turning point, lies on a vertical axis of symmetry: points equally far to either side have equal heights.

The $y$ -intercept is $(0,c)$ . The $x$ -intercepts come from $f(x)=0$ , so their horizontal coordinates are the equation's real roots. If there are two distinct roots $r,s$ , the symmetry line is halfway between them, $x=(r+s)/2$ . These graph results are not in the formula booklet: learn them.

Worked example 1: build a quadratic sketch

Question. Sketch $y=x^2-4x-5$ , labelling the intercepts and vertex.

Plan. Find the intercepts, use the midpoint of the roots for the symmetry line, then calculate the vertex's height.

\begin{aligned} x^2-4x-5&=(x-5)(x+1) && \text{factorise} \\ y=0&\Rightarrow x=-1\text{ or }5 && \text{zero product} \\ x=0&\Rightarrow y=-5 && y\text{-intercept} \\ x_{\mathrm{vertex}}&=\frac{-1+5}{2}=2 && \text{midpoint} \\ y_{\mathrm{vertex}}&=2^2-4(2)-5=-9 && \text{substitute} \end{aligned}

Why this step? Symmetry locates the vertex horizontally without needing an accurate plot of many points.

Answer. Draw a smooth upward parabola through $(-1,0)$ , $(5,0)$ and $(0,-5)$ , with minimum vertex $(2,-9)$ , as shown below.

Check. $f(1)=f(3)=-8$ , confirming symmetry about $x=2$ .

The picture joins the features with a curve, not straight segments. A sketch need not use an exact scale, but the relative positions and opening direction must agree with the algebra.

Upward parabola with roots at minus one and five, y-intercept minus five and vertex at two minus nine

The next example changes only the sign of the entire function. Each vertical coordinate changes sign, while the roots and symmetry line stay fixed.

Worked example 2: reverse the opening direction

Question. Sketch $y=-x^2+4x+5$ , labelling the intercepts and vertex.

Plan. Factorise and find the same key features, now checking for a maximum.

\begin{aligned} -x^2+4x+5&=-(x-5)(x+1) && \text{factorise} \\ y=0&\Rightarrow x=-1\text{ or }5 && \text{roots unchanged} \\ x=0&\Rightarrow y=5 && y\text{-intercept} \\ x_{\mathrm{vertex}}&=2 && \text{midpoint of roots} \\ y_{\mathrm{vertex}}&=-2^2+4(2)+5=9 && \text{substitute} \end{aligned}

Why this step? The negative squared-term coefficient changes the minimum into a maximum.

Answer. A smooth downward parabola has intercepts $(-1,0)$ , $(5,0)$ , $(0,5)$ and maximum vertex $(2,9)$ .

Check. Its values are the negatives of those in example 1; in particular, $y=8$ at both $x=1$ and $x=3$ .

The second sketch reflects the first in the horizontal axis; each labelled point agrees with the new equation.

Downward parabola with roots at minus one and five, y-intercept five and maximum vertex at two nine

AQA's sketch instruction asks for important features and a consistent shape. A list of coordinates is useful preparation, but does not itself draw the requested curve.

Watch out

Wrong: the $y$ -intercept of $y=-x^2+4x+5$ is $(5,0)$ . Right: put $x=0$ , giving $(0,5)$ . Coordinate order tells you which axis you are on.

Now locate features for a different quadratic. This check asks for the values rather than a drawing.

Completing the square and turning points

Completing the square makes the vertex visible even when the graph has no real roots to use as landmarks.

You already know

A real square is at least zero, and $(x-h)^2=0$ exactly when $x=h$ . The sign of the squared-term coefficient determines whether the vertex is a minimum or maximum.

Expand $(x+3)^2=x^2+6x+9$ . To reproduce $x^2+6x$ , subtract the extra $9$ . In general, halving the linear coefficient produces the required bracket:

x^2+bx+c=\left(x+\frac b2\right)^2+c-\frac{b^2}{4}

If the leading coefficient is not one, factor it from the quadratic and linear terms first.

Completed-square form

For real coefficients with $a\ne0$ ,

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

In the form $a(x-h)^2+v$ , the vertex is $(h,v)$ and symmetry line is $x=h$ . The extreme value is $v$ : a minimum if $a>0$ , a maximum if $a<0$ .

Not in the formula booklet: learn it.

Use the expansion to keep track of the constant inside the factored bracket. It will be multiplied by the leading coefficient too.

Worked example 3: expose a minimum

Question. Express $2x^2+12x+7$ in completed-square form and find its minimum.

Plan. Factor $2$ from the first two terms, complete the square, then use its least possible value.

\begin{aligned} 2x^2+12x+7&=2(x^2+6x)+7 && \text{factor the first terms} \\ &=2\bigl((x+3)^2-9\bigr)+7 && \text{complete the square} \\ &=2(x+3)^2-11 && \text{collect constants} \end{aligned}

Why this step? The compensating $-9$ is inside the bracket, so it contributes $-18$ , not $-9$ .

Answer. $2(x+3)^2-11$ , with minimum $-11$ at $x=-3$ , giving vertex $(-3,-11)$ .

Check. Expanding returns $2x^2+12x+18-11=2x^2+12x+7$ .

The next example changes the sign of the leading coefficient, retaining the same bracket after factorisation. The square now contributes a non-positive term.

Worked example 4: expose a maximum

Question. Express $-2x^2-12x+7$ in completed-square form and find its maximum.

Plan. Factor $-2$ , complete the same square and identify when the negative square is largest.

\begin{aligned} -2x^2-12x+7&=-2(x^2+6x)+7 && \text{factor }-2 \\ &=-2\bigl((x+3)^2-9\bigr)+7 && \text{complete the square} \\ &=-2(x+3)^2+25 && \text{collect constants} \end{aligned}

Why this step? $-2(x+3)^2\le0$ , so its largest value is zero, rather than an arbitrarily large positive number.

Answer. $-2(x+3)^2+25$ , with maximum $25$ at $x=-3$ , giving vertex $(-3,25)$ .

Check. Expansion gives $-2x^2-12x-18+25=-2x^2-12x+7$ .

Completed-square form also gives the full set of output values: $y\ge-11$ in example 3 and $y\le25$ in example 4. The bracket can become arbitrarily large in magnitude as $x$ varies.

Watch out

Wrong: the vertex of $2(x+3)^2-11$ is $(3,-11)$ . Right: the square vanishes when $x+3=0$ , so its vertex is $(-3,-11)$ .

A calculator can find a numerical vertex in an ordinary Find task. When the question requests completed-square form, give that form explicitly.

The discriminant and real roots

The discriminant counts real roots without requiring you to calculate them individually.

For the opening family, write $x^2-4x+c=(x-2)^2+c-4$ . Raising $c$ lifts the vertex without changing its horizontal position. At $c=4$ , both roots meet at the vertex. A tiny further upward shift removes both real roots at once.

Discriminant

For $ax^2+bx+c=0$ , where $a,b,c$ are real and $a\ne0$ , the discriminant is $D=b^2-4ac$ . A repeated root is one distinct real value occurring twice in the factorisation, as in $(x-2)^2=0$ .

The square root in the quadratic formula explains the three cases.

Discriminant test

The quadratic formula is

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

For real coefficients and $a\ne0$ , $D>0$ gives two distinct real roots, $D=0$ a repeated real root, and $D<0$ no real roots. A repeated root is $-b/(2a)$ .

Not in the formula booklet: learn it.

A positive square root gives two different choices of sign; a zero square root makes them equal; a negative radicand has no real square root.

Three upward parabolas showing two crossings, tangency and no crossing as the discriminant changes

This is also a consequence of completing the square, rather than a separate trick:

\begin{aligned} f(x)&=a\left(x+\frac b{2a}\right)^2-\frac D{4a} \\ f(x)=0&\Longleftrightarrow\left(x+\frac b{2a}\right)^2=\frac D{4a^2} \end{aligned}

Since $4a^2>0$ , the right side has the sign of $D$ . A real square can equal a positive number in two ways, zero in one way, and a negative number in no way. This derived identity is not in the formula booklet; derive it from completed-square form when needed.

Worked example 5: no real roots

Question. Determine the number of real roots of $3x^2-6x+4=0$ .

Plan. Evaluate the discriminant and interpret its sign.

\begin{aligned} a&=3,\quad b=-6,\quad c=4 && \text{identify coefficients} \\ D&=(-6)^2-4(3)(4) && \text{substitute with brackets} \\ &=36-48=-12<0 && \text{compare with zero} \end{aligned}

Why this step? A negative discriminant rules out real roots without needing an approximation to either root.

Answer. There are no real roots.

Check. $3x^2-6x+4=3(x-1)^2+1\ge1$ , so the graph stays above the axis.

Change just the constant from $4$ to $3$ . The vertex reaches the axis, producing a repeated root.

Worked example 6: a repeated root

Question. Determine the number of distinct real roots of $3x^2-6x+3=0$ , and find the root.

Plan. Check the discriminant, then use the zero square to identify the root.

\begin{aligned} D&=(-6)^2-4(3)(3) && \text{same test} \\ &=36-36=0 && \text{repeated-root case} \\ 3x^2-6x+3&=3(x-1)^2 && \text{factor the square} \\ 3(x-1)^2=0&\Rightarrow x=1 && \text{square must vanish} \end{aligned}

Why this step? Zero discriminant means the two formula signs give the same value, not two different intercepts.

Answer. One distinct real root, $x=1$ , repeated twice.

Check. Substitution gives $3-6+3=0$ , and the square form shows the graph touches the axis there.

No real roots does not by itself mean a positive function. For $D<0$ , an upward parabola lies above the axis and a downward one lies below it: their extreme value is $-D/(4a)$ .

Watch out

Wrong: “ $D=0$ , so there are no roots.” Right: there is a repeated real root. “No real roots” requires $D<0$ .

Change the constant again to move the graph below its tangency position.

Root conditions involving parameters

When coefficients contain a parameter, the requested roots become a condition on that parameter.

Translate the words first: two distinct real roots need $D>0$ ; a repeated root needs $D=0$ ; real roots, allowing equality, need $D\ge0$ . These conditions assume the squared-term coefficient is nonzero. None of these parameter rules is in the formula booklet: use the discriminant test you have learned.

Worked example 7: find all parameter cases

Question. For real $k$ , classify the real roots of $x^2+2kx+k+2=0$ .

Plan. Factorise the discriminant in $k$ , then consider the signs between its zeros.

\begin{aligned} D&=(2k)^2-4(k+2) && a=1\ne0 \\ &=4(k^2-k-2) && \text{collect terms} \\ &=4(k-2)(k+1) && \text{factorise} \end{aligned}

The zeros are $k=-1,2$ . If $k<-1$ , both factors are negative and the product is positive. If $-1<k<2$ , one is negative and one positive. If $k>2$ , both are positive.

Why this step? The discriminant itself is a quadratic in the parameter; its sign changes at its roots, so solving only $D=0$ would find boundaries without the intervals.

Answer. Two distinct real roots for $k<-1$ or $k>2$ ; no real roots for $-1<k<2$ ; a repeated root for $k=-1$ or $k=2$ . The repeated root is $x=-k$ : respectively $1$ and $-2$ .

Check. At $k=0$ , the equation is $x^2+2=0$ , with no real roots. At the boundaries it becomes $(x-1)^2=0$ and $(x+2)^2=0$ .

The next example changes which coefficient contains the parameter: it is now the leading coefficient. Before using the discriminant, set aside the value where that coefficient vanishes.

Worked example 8: a disappearing squared term

Question. Find all real $p$ for which $(p-1)x^2+2x+1=0$ has at least one real solution.

Plan. Treat the quadratic case $p\ne1$ , then check $p=1$ in the original equation.

\begin{aligned} D&=2^2-4(p-1)(1)=8-4p && p\ne1 \\ D\ge0&\Longleftrightarrow p\le2 && \text{allow repeated roots} \\ p=1&\Rightarrow 2x+1=0 && \text{original equation} \\ &\Rightarrow x=-\frac12 && \text{a real linear solution} \end{aligned}

Why this step? The discriminant classifies quadratic roots; it does not classify the linear equation left when $p=1$ .

Answer. $p\le2$ , including the separate linear case $p=1$ .

Check. At $p=2$ , $(x+1)^2=0$ has a repeated root; at $p=3$ , $D=-4$ , so there is no real solution.

A boundary can be included because a repeated root is allowed, or because a separate linear equation has a real solution. Give the reason: the final interval alone can hide a missed exceptional case.

Watch out

Wrong: “Real roots means $D>0$ .” Right: $D\ge0$ unless the question says distinct. Also join the two outer intervals in example 7 with “or”; no value can be less than $-1$ and greater than $2$ at once.

The opening family is now explained: changing $c$ in $x^2-4x+c$ changes $D=16-4c$ . Its sign changes at $c=4$ , exactly when the vertex reaches the axis.

Quick Recap

Intercepts, opening direction and symmetry determine the essential quadratic sketch.
Completed-square form gives the vertex and explains the extreme value.
Positive, zero and negative discriminants give two distinct real roots, a repeated real root and no real roots.
Translate the exact root condition, solve its parameter inequality and check any vanishing leading coefficient separately.

Changing the opening family's constant raises the vertex: $c<4$ gives two crossings, $c=4$ one tangency, and $c>4$ none.

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B3 - Quadratic graphs and discriminants

Features of a quadratic graph

Completing the square and turning points

The discriminant and real roots

Root conditions involving parameters

Quick Recap

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