2. Quadratic functions

Learning objectives

2.1Find maximum and minimum values• Find the maximum or minimum value of the quadratic function f: x -> ax^2 + bx + c by completing the square or by differentiation.2.2Use vertex values to sketch or find range• Use the maximum or minimum value of f(x) to sketch y = f(x) or determine the range for a given domain.• Use correct notation to write a domain or range.2.3Use discriminant conditions• Know the conditions for f(x) = 0 to have two real roots, two equal roots or no real roots.• Relate the discriminant to the roots of the equation.• Apply the related conditions for a line to intersect a curve, be tangent to a curve or not intersect a curve.2.4Solve quadratic equations• Solve quadratic equations for real roots.• Use factorisation, the quadratic formula and completing the square.• The quadratic formula is given in the List of formulas.• On the calculator paper, correct answers are acceptable without working.2.5Solve quadratic inequalities• Find the solution set for quadratic inequalities graphically or algebraically.• Write solutions in the correct form, such as -3 < x < 4 or x < 1 or x > 6.

For $f(x)=ax^2+bx+c$ with $a\ne0$ , the turning point gives the quadratic's minimum when $a>0$ and maximum when $a<0$ . Its $x$ -coordinate is the axis of symmetry, $x=-b/(2a)$ .

Completing the square writes the function as $f(x)=a(x-h)^2+k$ . Because $(x-h)^2\geq0$ , the turning point is $(h,k)$ : $k$ is the minimum if $a>0$ and the maximum if $a<0$ .

$2x^2+3x-4=2\left(x+\frac34\right)^2-\frac{41}{8}\quad\Rightarrow\quad\min f=-\frac{41}{8}\text{ at }x=-\frac34$

Alternatively, differentiate: $f'(x)=2ax+b$ . Set $f'(x)=0$ to obtain $x=-b/(2a)$ , substitute this value into $f$ , then use the sign of $a$ to classify the turning point. Both methods locate the same point.

Give both the extreme value and where it occurs when asked. The value is the $y$ -coordinate; it is not the same as the $x$ -coordinate of the turning point.

Vertex form $f(x)=a(x-h)^2+k$ reveals the turning point $(h,k)$ , axis $x=h$ , and opening direction. These features fix the basic sketch and the unrestricted range.

Situation	Range
$a>0$ , domain $\mathbb R$	$f(x)\geq k$
$a<0$ , domain $\mathbb R$	$f(x)\leq k$
restricted domain	compare the vertex, included endpoints and end behaviour that lie in the domain

For the sketch, mark the vertex and axis of symmetry, then find any real $x$ -intercepts and the $y$ -intercept. Draw a smooth symmetric parabola opening upward for $a>0$ or downward for $a<0$ .

A restricted domain can remove the vertex. For $f(x)=(x-2)^2-5$ with domain $x\geq4$ , the permitted branch starts at $(4,-1)$ and rises, so the range is $f(x)\geq-1$ , not $f(x)\geq-5$ .

A domain restricts $x$ ; a range restricts $f(x)$ or $y$ . Check whether an endpoint is included before choosing $\leq$ or $<$ .

For $ax^2+bx+c=0$ , the discriminant $D=b^2-4ac$ tells how many real roots exist without solving the equation.

Discriminant	Roots of the equation	Line and curve
$D>0$	two distinct real roots	two intersections
$D=0$	two equal real roots	tangent; one point of contact
$D<0$	no real roots	no intersection

For a line and a curve, first equate their $y$ -expressions and rearrange to one quadratic in $x$ . Only then identify $a$ , $b$ and $c$ and apply the required condition to $D$ . If a parameter is present, solving the resulting equation or inequality gives its allowed values.

$y=x^2+1,\ y=kx\quad\Rightarrow\quad x^2-kx+1=0,\quad D=k^2-4$

Thus the line cuts the curve twice when $|k|>2$ , is tangent when $k=\pm2$ , and does not meet it when $|k|<2$ .

Use $D\geq0$ for ‘real roots’, but $D>0$ for ‘two different real roots’. Tangency requires equality, not merely a non-negative discriminant.

First rearrange the equation to $ax^2+bx+c=0$ . Then choose the method that exposes the roots with the least unnecessary work.

Method	Best choice when	Essential move
factorisation	factors are visible	use the zero-product rule
quadratic formula	factors are awkward	substitute $a,b,c$ with their signs
completing the square	vertex form is useful	isolate the square, then square-root both sides

$x^2-5x+4=0\Rightarrow(x-1)(x-4)=0\Rightarrow x=1\text{ or }x=4$

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

For $x^2-6x+1=0$ , write $(x-3)^2=8$ , so $x=3\pm2\sqrt2$ . The $\pm$ is required because both numbers have the same square.

Substitute each proposed root into the original equation. If $D<0$ , there are no real roots in this syllabus context; do not replace $\sqrt{D}$ with a real number.

A quadratic inequality asks where the graph lies above, on or below the $x$ -axis. Its real roots are critical values where the sign may change.

Rearrange so one side is zero, solve the corresponding quadratic equation, then split the number line at its real roots. Determine the sign in each interval by a test value or by the parabola's opening direction. Include a root only for $\leq$ or $\geq$ .

$(x+2)(x-3)\leq0\quad\Rightarrow\quad-2\leq x\leq3$

For $x^2-x-6\geq0$ , the roots are $-2$ and $3$ and the upward-opening graph is on or above the axis outside them. Hence $x\leq-2$ or $x\geq3$ .

Use ‘and’ for one interval such as $-2<x<3$ , but ‘or’ for two separate outer intervals. If there are no real roots, the quadratic keeps the sign of its leading coefficient for every real $x$ .

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2. Quadratic functions

Learning objectives

Find the turning value of a quadratic

Use the turning point to sketch and find a range

Use the discriminant to count roots and intersections

Choose a method to solve a quadratic equation

Solve a quadratic inequality by interval signs

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