A quadratic function is any function where the highest power of $x$ is 2. The standard form is:
where $a \neq 0$, and $a$, $b$, $c$ are constants
Its graph is a smooth, symmetrical curve called a parabola.
When $a > 0$ (positive)
The parabola is U-shaped — it opens upward and has a minimum turning point.
When $a < 0$ (negative)
The parabola is ∩-shaped — it opens downward and has a maximum turning point.
Key features on graph
Every parabola has: a y-intercept, possibly 0, 1 or 2 roots, and exactly one turning point.
Key features of a quadratic graph: roots (red), y-intercept (orange), vertex (purple)
No real roots
Graph doesn't cross x-axis
e.g. $y = x^2 + 2$
One root (repeated)
Graph touches x-axis
e.g. $y = x^2$
Two roots
Graph crosses x-axis twice
e.g. $y = x^2 - 4$
Method 1: Finding Roots by Factorising
- Set $y = 0$ to get the equation $ax^2 + bx + c = 0$
- Factorise the quadratic into two brackets
- Set each bracket equal to zero and solve for $x$
Method 2: Finding Roots using the Quadratic Formula
When a quadratic is hard to factorise, use the quadratic formula:
The expression $b^2 - 4ac$ is called the discriminant — it tells you how many roots there are.
| Discriminant $b^2 - 4ac$ | Number of roots | Graph behaviour |
|---|---|---|
| $b^2 - 4ac > 0$ | Two distinct real roots | Crosses x-axis at two points |
| $b^2 - 4ac = 0$ | One repeated root | Touches x-axis at one point |
| $b^2 - 4ac < 0$ | No real roots | Doesn't touch x-axis |
Verify: substitute $x = 0$: $y = 3(0) - 7(0) + 4 = 4$ ✓
Substitute this $x$-value back into the equation to find the $y$-coordinate of the turning point.
Finding the Turning Point by Completing the Square
Completing the square rewrites $y = ax^2 + bx + c$ in the form $y = a(x + p)^2 + q$, which directly reveals the turning point at $(-p,\ q)$.
- Start with $y = x^2 + bx + c$
- Halve the coefficient of $x$ to get $\dfrac{b}{2}$, then write $(x + \dfrac{b}{2})^2$
- Subtract $\left(\dfrac{b}{2}\right)^2$ to compensate for the extra term introduced: $$y = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c$$
- Simplify the constants — this gives you $a(x+p)^2 + q$
- The turning point is $(-p,\ q)$
In the exam you may be given a graph and asked to read off its features. Here's what to look for:
| Feature | Where to look | What it looks like |
|---|---|---|
| Roots | Where the curve crosses the x-axis | Points of the form $(x, 0)$ |
| y-intercept | Where the curve crosses the y-axis | Point of the form $(0, y)$ |
| Turning point | The lowest (U-shape) or highest (∩-shape) point of the curve | The tip of the parabola |
| Axis of symmetry | The vertical line through the turning point | Dashed vertical line $x = k$ |
| Values of $y$ | Read across from the y-axis at a given $x$ | Read the height of the curve |
| Solutions to $f(x)=k$ | Draw horizontal line $y=k$, find where it meets curve | x-coordinates at intersection |
Using the annotated diagram from Section 1:
To sketch $y = ax^2 + bx + c$ fully, find and plot all key features:
- Shape: If $a > 0$, U-shape. If $a < 0$, ∩-shape.
- y-intercept: Always at $(0, c)$.
- Roots: Solve $ax^2 + bx + c = 0$ by factorising or the quadratic formula.
- Turning point: Use $x = -\dfrac{b}{2a}$ then substitute to find $y$; or use completed square form.
- Axis of symmetry: The vertical line $x = -\dfrac{b}{2a}$.
- Plot all points, then draw a smooth symmetrical curve through them.
Standard form
$y = ax^2 + bx + c$ — $a \ne 0$
$a>0$: U-shape (min) $a<0$: ∩-shape (max)
y-intercept
Always $(0,\ c)$ — just read the constant term.
Roots
Set $y=0$, then factorise or use $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$
Discriminant
$b^2-4ac>0$: 2 roots
$b^2-4ac=0$: 1 root
$b^2-4ac<0$: no real roots
Axis of symmetry
$x = -\dfrac{b}{2a}$
Or midpoint of roots: $\dfrac{x_1+x_2}{2}$
Turning point
$x = -\dfrac{b}{2a}$, then sub back. Or complete the square: $a(x+p)^2+q$ → vertex $(-p,\ q)$.
Completed square
$y = (x+p)^2 + q$ → vertex at $(-p,\ q)$. Set $y=0$ to find roots from this form.
Symmetry shortcut
The turning point's x-coord is the average of the two roots: $x = \dfrac{x_1 + x_2}{2}$
For the quadratic $y = x^2 - 3x - 10$, find the roots by factorising. State the coordinates where the graph crosses the x-axis.
▶ Show solution
Set $y = 0$: $\; x^2 - 3x - 10 = 0$
Find two numbers multiplying to $-10$ and adding to $-3$: $+2$ and $-5$.
$(x + 2)(x - 5) = 0$
$x = -2$ or $x = 5$
The graph crosses the x-axis at $(-2,\ 0)$ and $(5,\ 0)$.
Write down the y-intercept of each of the following:
(a) $y = x^2 - 5x + 7$ (b) $y = 3x^2 + 2x - 8$ (c) $y = -2x^2 + x$
▶ Show solution
(a) $c = 7$, so y-intercept is $\mathbf{(0,\ 7)}$
(b) $c = -8$, so y-intercept is $\mathbf{(0,\ -8)}$
(c) There is no constant term, so $c = 0$; y-intercept is $\mathbf{(0,\ 0)}$ — the curve passes through the origin.
Find the turning point and axis of symmetry of $y = x^2 + 6x + 8$. State whether it is a maximum or minimum.
▶ Show solution
$a = 1,\; b = 6,\; c = 8$
Axis of symmetry: $x = -\dfrac{6}{2(1)} = -3$
Substitute: $y = (-3)^2 + 6(-3) + 8 = 9 - 18 + 8 = -1$
Turning point: $(-3,\; -1)$ — it is a minimum since $a > 0$.
Axis of symmetry: $x = -3$
Complete the square for $y = x^2 - 8x + 10$. Hence write down the coordinates of the turning point and the equation of the axis of symmetry.
▶ Show solution
Halve the coefficient of $x$: $\frac{-8}{2} = -4$
$y = (x - 4)^2 - 16 + 10$
$$y = (x - 4)^2 - 6$$
Turning point: $(4,\; -6)$ (minimum, since $a = 1 > 0$)
Axis of symmetry: $x = 4$
Use the quadratic formula to find the roots of $y = x^2 - 4x + 1$. Give your answers correct to 2 decimal places.
▶ Show solution
$a=1,\; b=-4,\; c=1$
Discriminant: $(-4)^2 - 4(1)(1) = 16 - 4 = 12$
$$x = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$$
$x = 2 + \sqrt{3} \approx 3.73$ or $x = 2 - \sqrt{3} \approx 0.27$
Roots: $x \approx 3.73$ and $x \approx 0.27$
Complete the square for $y = x^2 + 4x + 7$. How many real roots does this quadratic have? Explain your answer.
▶ Show solution
Halve the coefficient of $x$: $\frac{4}{2} = 2$
$y = (x + 2)^2 - 4 + 7 = (x+2)^2 + 3$
Since $(x+2)^2 \geq 0$ always, the minimum value of $y$ is $0 + 3 = 3$, which is above the x-axis.
The quadratic has no real roots — the graph never crosses the x-axis.
(Check: discriminant $= 4^2 - 4(1)(7) = 16 - 28 = -12 < 0$ ✓)
Sketch the graph of $y = x^2 - 2x - 8$. Clearly label the roots, y-intercept and turning point.
▶ Show solution
Shape: $a = 1 > 0$ → U-shaped (minimum)
y-intercept: $(0,\; -8)$
Roots: $x^2 - 2x - 8 = 0 \Rightarrow (x-4)(x+2)=0 \Rightarrow x=4$ or $x=-2$
So roots at $(-2,\; 0)$ and $(4,\; 0)$
Turning point: $x = -\dfrac{-2}{2} = 1$; $y = 1 - 2 - 8 = -9$ → minimum at $(1,\; -9)$
Axis of symmetry: $x=1$ (check: $\frac{-2+4}{2}=1$ ✓)
Draw a U-shaped curve through $(-2,0)$, $(0,-8)$, $(1,-9)$, $(4,0)$, symmetric about $x=1$.
The graph of $y = -x^2 + 6x - 5$ has a maximum turning point.
(a) Find the roots. (b) Find the maximum turning point. (c) Write the y-intercept.
▶ Show solution
(a) Roots: Set $y=0$: $-x^2+6x-5=0 \Rightarrow x^2-6x+5=0 \Rightarrow (x-1)(x-5)=0$
Roots: $x = 1$ and $x = 5$ → graph crosses x-axis at $(1,0)$ and $(5,0)$
(b) Turning point: $a=-1,\;b=6$; axis: $x = -\dfrac{6}{2(-1)} = 3$
$y = -(3)^2 + 6(3) - 5 = -9 + 18 - 5 = 4$ → Maximum at $(3,\;4)$
(c) y-intercept: $(0,\;-5)$
Complete the square for $y = 2x^2 - 12x + 13$. Hence find the minimum value of $y$ and the value of $x$ at which it occurs.
▶ Show solution
Factor out 2 from the $x^2$ and $x$ terms:
$y = 2(x^2 - 6x) + 13$
Complete the square inside: $x^2 - 6x = (x-3)^2 - 9$
$y = 2\big[(x-3)^2 - 9\big] + 13 = 2(x-3)^2 - 18 + 13$
$$y = 2(x-3)^2 - 5$$
Since $2(x-3)^2 \geq 0$, minimum occurs when $(x-3)^2 = 0$, i.e. at $x=3$.
Minimum value of $y = -5$, occurring at $x = 3$.
Turning point: $(3,\; -5)$
The quadratic $y = x^2 - 6x + k$ has exactly one real root.
(a) Use the discriminant to find the value of $k$.
(b) Find the coordinates of the turning point.
(c) Write the equation of the axis of symmetry.
▶ Show solution
(a) For exactly one root: discriminant $= 0$
$b^2 - 4ac = 0 \Rightarrow (-6)^2 - 4(1)(k) = 0 \Rightarrow 36 - 4k = 0 \Rightarrow k = 9$
$k = 9$, so the equation is $y = x^2 - 6x + 9 = (x-3)^2$
(b) From $(x-3)^2$, the turning point is $(3,\ 0)$. (The parabola just touches the x-axis at its vertex.)
(c) Axis of symmetry: $x = 3$