Quadratic Functions – A Level Maths Year 1

1 The Graph of a Quadratic

A quadratic function has the form $\mathrm{f}(x) = ax^2 + bx + c$ with $a \neq 0$. Its graph is a parabola: a symmetrical curve with a single turning point (the vertex).

Feature	How to find it
Shape	$a > 0$: $\cup$-shaped, with a minimum. $a < 0$: $\cap$-shaped, with a maximum.
$y$-intercept	Put $x = 0$: the curve crosses the $y$-axis at $(0,\, c)$.
Roots ($x$-intercepts)	Solve $ax^2 + bx + c = 0$.
Line of symmetry	$x = -\dfrac{b}{2a}$, halfway between the roots.
Vertex	Complete the square (Section 3).

Worked Example 1 — Sketching a quadratic

Sketch $y = x^2 - 2x - 3$, labelling the intercepts and the turning point.

①$a = 1 > 0$, so the curve is $\cup$-shaped.

②$x = 0$ gives $y = -3$: the $y$-intercept is $(0,\,-3)$.

③$x^2 - 2x - 3 = (x+1)(x-3) = 0$ gives roots $x = -1$ and $x = 3$.

④The line of symmetry is halfway between the roots: $x = 1$. Then $y = 1 - 2 - 3 = -4$, so the vertex is $(1,\,-4)$.

What a sketch needs: the correct shape, and every point where the curve meets an axis, labelled with its coordinates. The turning point should be labelled if you know it. It does not need to be drawn to scale.

2 Solving Quadratic Equations

The roots of $\mathrm{f}(x) = ax^2 + bx + c$ are the solutions of $\mathrm{f}(x) = 0$. There are three methods, and you should be fluent in all of them.

The quadratic formula

$ax^2 + bx + c = 0 \;\Rightarrow\; x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Method	Use it when…
Factorising	The quadratic factorises neatly — always check first.
Quadratic formula	It does not factorise, or you need exact surd answers.
Completing the square	The question asks for it, or you also need the vertex.

Worked Example 2 — Using the formula

Solve $3x^2 - 5x - 1 = 0$, giving your answers in exact form and to $3$ significant figures.

①$a = 3$, $b = -5$, $c = -1$.

②$b^2 - 4ac = 25 - 4(3)(-1) = 25 + 12 = 37$

③$x = \dfrac{5 \pm \sqrt{37}}{6}$

④$x = 1.85$ or $x = -0.180$ (3 s.f.)

Careful with $-b$ when $b$ is negative: $-(-5) = +5$. Writing $\dfrac{-5 \pm \sqrt{37}}{6}$ is the most common slip in this topic.

Calculators: your calculator's equation solver is a good way to check roots. But if a question says "show detailed reasoning" or "give exact values", you must write out the factorising, the formula or the completed square.

3 Completing the Square

The completed-square form

$x^2 + bx + c \equiv \left(x + \dfrac{b}{2}\right)^2 - \left(\dfrac{b}{2}\right)^2 + c$
$ax^2 + bx + c \equiv a(x + p)^2 + q$

Why it is so useful. In $a(x+p)^2 + q$ the bracket is squared, so it is never negative and equals zero only when $x = -p$. So the vertex is at $(-p,\, q)$, and $q$ is the minimum value (if $a > 0$) or maximum value (if $a < 0$).

Worked Example 3 — Leading coefficient not 1

Write $2x^2 - 12x + 7$ in the form $a(x+p)^2 + q$, and state the coordinates of the turning point of $y = 2x^2 - 12x + 7$.

①Take out the factor $2$ from the $x$ terms only: $2\left(x^2 - 6x\right) + 7$

②Complete the square inside: $x^2 - 6x \equiv (x-3)^2 - 9$

③$2\left[(x-3)^2 - 9\right] + 7 = 2(x-3)^2 - 18 + 7$

④$2x^2 - 12x + 7 \equiv 2(x-3)^2 - 11$

⑤The minimum point is $(3,\,-11)$.

The $-9$ is inside the bracket that gets multiplied by $2$. Forgetting to multiply it — writing $2(x-3)^2 - 9 + 7$ — is the classic error.

Worked Example 4 — A negative leading coefficient

Express $1 + 4x - x^2$ in the form $q - (x+p)^2$ and hence find the maximum value of the expression.

①$1 + 4x - x^2 = -\left(x^2 - 4x\right) + 1$

②$= -\left[(x-2)^2 - 4\right] + 1 = -(x-2)^2 + 4 + 1$

③$1 + 4x - x^2 \equiv 5 - (x-2)^2$

④$(x-2)^2 \geqslant 0$, so the expression is at most $5$. The maximum value is $5$, when $x = 2$.

Check by expanding. $5 - (x^2 - 4x + 4) = 1 + 4x - x^2$ ✓. It takes ten seconds and catches nearly every sign error. Completing the square is also exactly how the quadratic formula is proved.

4 The Discriminant

The expression under the square root in the formula, $b^2 - 4ac$, is called the discriminant. Its sign tells you how many real roots there are without solving anything.

The discriminant $\Delta = b^2 - 4ac$

$b^2 - 4ac > 0$: two distinct real roots
$b^2 - 4ac = 0$: one repeated root (the graph touches the $x$-axis)
$b^2 - 4ac < 0$: no real roots (the graph misses the $x$-axis)

Worked Example 5 — Equal roots

The equation $kx^2 + 4x + (k - 3) = 0$ has equal roots. Find the possible values of $k$.

①Equal roots means $b^2 - 4ac = 0$, with $a = k$, $b = 4$, $c = k-3$.

②$16 - 4k(k-3) = 0 \;\Rightarrow\; 16 - 4k^2 + 12k = 0$

③Divide by $-4$: $k^2 - 3k - 4 = 0$

④$(k-4)(k+1) = 0$, so $k = 4$ or $k = -1$.

Check: $k = 4$ gives $4x^2 + 4x + 1 = (2x+1)^2$ ✓, and $k = -1$ gives $-x^2 + 4x - 4 = -(x-2)^2$ ✓. Both are perfect squares, as they must be.

Worked Example 6 — No real roots

Find the set of values of $k$ for which $x^2 + kx + 9 = 0$ has no real roots.

①No real roots: $b^2 - 4ac < 0$, so $k^2 - 36 < 0$.

②$(k-6)(k+6) < 0$

③The graph of $k^2 - 36$ is $\cup$-shaped and negative between its roots: $-6 < k < 6$.

Discriminant questions very often finish with a quadratic inequality. "Two distinct roots" means $> 0$; "real roots" (with no "distinct") means $\geqslant 0$.

If the $x^2$ coefficient contains $k$, check whether it can be zero. When $a = 0$ the equation is linear, not quadratic, and the discriminant tells you nothing.

5 Quadratics in Disguise

Some equations are not quadratics in $x$ but are quadratics in some other expression, such as $x^2$, $\sqrt{x}$ or $x^{1/3}$. Spot the pattern "something squared, the same something, a constant" and substitute.

Spotting the hidden quadratic

$x^4 = \left(x^2\right)^2$ · $x = \left(x^{1/2}\right)^2$ · $x^{2/3} = \left(x^{1/3}\right)^2$
Let $u$ = the "something", solve for $u$, then convert back to $x$

Worked Example 7 — A quadratic in $x^2$

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

①Let $u = x^2$. Then $u^2 - 5u + 4 = 0$.

②$(u-1)(u-4) = 0$, so $u = 1$ or $u = 4$.

③$x^2 = 1 \Rightarrow x = \pm 1$; $x^2 = 4 \Rightarrow x = \pm 2$.

④Four solutions: $x = -2, -1, 1, 2$.

Worked Example 8 — A quadratic in $\sqrt{x}$

Solve $x + \sqrt{x} - 6 = 0$.

①Let $u = \sqrt{x} = x^{1/2}$, so $x = u^2$: $u^2 + u - 6 = 0$.

②$(u+3)(u-2) = 0$, so $u = -3$ or $u = 2$.

③$\sqrt{x}$ is never negative, so reject $u = -3$.

④$\sqrt{x} = 2 \Rightarrow x = 4$. Check: $4 + 2 - 6 = 0$ ✓

Always go back and check whether each value of $u$ is possible. $\sqrt{x} = -3$ has no solution, and squaring it to get $x = 9$ gives a false root: $9 + 3 - 6 = 6 \neq 0$.

Worked Example 9 — A quadratic in $x^{1/3}$

Solve $x^{2/3} - 3x^{1/3} - 4 = 0$.

①Let $u = x^{1/3}$: $u^2 - 3u - 4 = 0$.

②$(u-4)(u+1) = 0$, so $u = 4$ or $u = -1$.

③Cube both: $x = 4^3 = 64$ or $x = (-1)^3 = -1$.

Unlike a square root, a cube root can be negative, so both answers stand. See Indices and Surds for fractional powers.

6 Quick Reference

Shape

$a > 0$: $\cup$, minimum. $a < 0$: $\cap$, maximum.

$y$-intercept

$(0,\,c)$.

Symmetry

$x = -\dfrac{b}{2a}$, midway between the roots.

Formula

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

Completed square

$a(x+p)^2 + q$: vertex $(-p,\,q)$.

Factor out $a$

From the $x$ terms only, then multiply back carefully.

Discriminant

$>0$ two roots, $=0$ one repeated, $<0$ none.

"Real roots"

$b^2 - 4ac \geqslant 0$; "distinct" means $>0$.

Hidden quadratic

Substitute $u = x^2$, $\sqrt{x}$, $x^{1/3}$, …

False roots

$\sqrt{x}$ cannot be negative — reject.

7 Practice Questions

Question 1

Solve $2x^2 + 5x - 12 = 0$.

▶ Show solution

$ac = -24$: two numbers multiplying to $-24$ and adding to $5$ are $8$ and $-3$.

$2x^2 + 8x - 3x - 12 = 2x(x+4) - 3(x+4) = (2x-3)(x+4)$

$x = \dfrac{3}{2}$ or $x = -4$.

Question 2

Solve $x^2 + 6x + 4 = 0$, giving your answers in the form $p \pm \sqrt{q}$.

▶ Show solution

Completing the square: $(x+3)^2 - 9 + 4 = 0$, so $(x+3)^2 = 5$.

$x + 3 = \pm\sqrt{5}$, so $x = -3 \pm \sqrt{5}$.

(The formula gives $\dfrac{-6 \pm \sqrt{20}}{2} = \dfrac{-6 \pm 2\sqrt{5}}{2} = -3 \pm \sqrt{5}$, the same.)

Question 3

Write $x^2 + 8x + 3$ in the form $(x+p)^2 + q$ and hence write down the coordinates of the turning point of $y = x^2 + 8x + 3$.

▶ Show solution

$x^2 + 8x + 3 = (x+4)^2 - 16 + 3 = (x+4)^2 - 13$

Turning point (a minimum): $(-4,\,-13)$.

Question 4

(a) Express $3x^2 - 6x + 5$ in the form $a(x+p)^2 + q$. (b) Hence explain why $3x^2 - 6x + 5 > 0$ for all real $x$.

▶ Show solution

(a) $3\left(x^2 - 2x\right) + 5 = 3\left[(x-1)^2 - 1\right] + 5 = 3(x-1)^2 + 2$

(b) $(x-1)^2 \geqslant 0$ for all $x$, so $3(x-1)^2 \geqslant 0$ and $3(x-1)^2 + 2 \geqslant 2 > 0$.

Question 5

Find the value of the discriminant of $2x^2 - 3x + 4$, and hence state the number of real roots of $2x^2 - 3x + 4 = 0$.

▶ Show solution

$b^2 - 4ac = (-3)^2 - 4(2)(4) = 9 - 32 = -23$

The discriminant is negative, so there are no real roots. The graph of $y = 2x^2 - 3x + 4$ lies entirely above the $x$-axis.

Question 6

The equation $x^2 + (k+3)x + 4k = 0$ has equal roots. Find the possible values of $k$.

▶ Show solution

Equal roots: $(k+3)^2 - 4(1)(4k) = 0$

$k^2 + 6k + 9 - 16k = 0 \;\Rightarrow\; k^2 - 10k + 9 = 0$

$(k-1)(k-9) = 0$, so $k = 1$ or $k = 9$.

Check: $k = 1$ gives $x^2 + 4x + 4 = (x+2)^2$ ✓; $k = 9$ gives $x^2 + 12x + 36 = (x+6)^2$ ✓.

Question 7

Solve $x^4 - 10x^2 + 9 = 0$.

▶ Show solution

Let $u = x^2$: $u^2 - 10u + 9 = 0$, so $(u-1)(u-9) = 0$.

$x^2 = 1$ or $x^2 = 9$.

$x = -3,\; -1,\; 1,\; 3$.

Question 8

Solve $2x - 7\sqrt{x} + 3 = 0$.

▶ Show solution

Let $u = \sqrt{x}$, so $x = u^2$: $2u^2 - 7u + 3 = 0$.

$(2u - 1)(u - 3) = 0$, so $u = \tfrac12$ or $u = 3$.

Both are positive, so both are valid: $x = \left(\tfrac12\right)^2 = \tfrac14$ or $x = 3^2 = 9$.

Check $x = 9$: $18 - 21 + 3 = 0$ ✓. $x = \tfrac14$: $\tfrac12 - \tfrac72 + 3 = 0$ ✓.

Question 9

Find the set of values of $p$ for which the equation $px^2 + 4x + p = 0$ has two distinct real roots.

▶ Show solution

Two distinct real roots: $b^2 - 4ac > 0$, so $16 - 4p^2 > 0$.

$p^2 < 4$, giving $-2 < p < 2$.

But if $p = 0$ the equation is $4x = 0$, which is linear with only one root. So we must exclude it:

$-2 < p < 2$, $p \neq 0$ i.e. $\{p : -2 < p < 0\} \cup \{p : 0 < p < 2\}$.

Question 10

A ball is thrown upwards. Its height $h$ metres above the ground $t$ seconds later is modelled by $h = 1.5 + 14t - 4.9t^2$.

(a) By completing the square, find the maximum height of the ball and the time at which it occurs. (b) Use the discriminant to show that the ball never reaches a height of $12$ m. (c) Find the time at which the ball hits the ground, to $3$ significant figures. (d) State one limitation of the model.

▶ Show solution

(a) $h = -4.9\left(t^2 - \dfrac{20}{7}t\right) + 1.5$ (since $14 \div 4.9 = \tfrac{20}{7}$)

$= -4.9\left[\left(t - \dfrac{10}{7}\right)^2 - \dfrac{100}{49}\right] + 1.5 = -4.9\left(t - \dfrac{10}{7}\right)^2 + 10 + 1.5$

$h = 11.5 - 4.9\left(t - \dfrac{10}{7}\right)^2$

Maximum height $11.5$ m, at $t = \dfrac{10}{7} \approx 1.43$ s.

(b) $h = 12$ gives $4.9t^2 - 14t + 10.5 = 0$. Discriminant $= 14^2 - 4(4.9)(10.5) = 196 - 205.8 = -9.8 < 0$, so there is no real $t$: the ball never reaches $12$ m. (Consistent with (a): $12 > 11.5$.)

(c) $h = 0$: $4.9t^2 - 14t - 1.5 = 0$

$t = \dfrac{14 \pm \sqrt{196 + 29.4}}{9.8} = \dfrac{14 \pm \sqrt{225.4}}{9.8}$

The negative root is not valid ($t \geqslant 0$), so $t = \dfrac{14 + 15.013\ldots}{9.8} = 2.96$ s (3 s.f.).

(d) For example: it ignores air resistance; it treats the ball as a particle; it only applies until the ball hits the ground (for larger $t$ it gives negative heights).

Quadratic Functions · A Level Maths Year 1 · Created with MathJax