Understand · explore · practise
Understand function notation, sketch quadratic graphs, find intercepts and turning points, and work out the range from the allowed inputs.
Before you startSolving quadratics and completing the square
01 / Functions
A function is a rule that assigns exactly one output to each allowed input. The domain is the set of allowed inputs; the range is the set of outputs actually produced.
f(x) = x² − 2x − 3
f(4) = 4² − 2(4) − 3 = 5
f(4) means “the value of f at 4”. It does not mean f multiplied by 4. If the input is an expression, put that entire expression into every x position.
f(t + 1) = (t + 1)² − 2(t + 1) − 3
= t² − 4
x ∈ ℝ means x belongs to the real numbers. A polynomial is defined for all real x unless its domain is restricted. For g(x) = √(x − 2), real outputs require x ≥ 2, and its range is g(x) ≥ 0.
The roots of f are inputs for which f(x) = 0. The graph y = f(x) places each input-output pair at (x, f(x)).
f(x) = x² − 2x − 3
The rule still gives one output for every input.
f(a) = 5 ⇒ a² − 2a − 8 = 0
An unknown input is found by solving an equation.
(a − 4)(a + 2) = 0
Factorise.
a = 4 or a = −2
Both inputs produce 5. That does not stop f being a function.
02 / Sketch
A quadratic graph is a parabola. For y = ax² + bx + c with a ≠ 0, identify these features before drawing:
For f(x) = x² − 2x − 3 = (x − 1)² − 4, the turning point is (1, −4), the roots are −1 and 3, and the y-intercept is −3. The two roots are equally far from x = 1.
A sketch needs labelled axes and key coordinates; it need not be a point-by-point plot. In a restricted domain, draw only the allowed part.
For all real x, the minimum is −4 at x = 1. Range: y ≥ −4.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Domain & range
For all real inputs, a(x − h)² + k has range y ≥ k when a > 0, and y ≤ k when a < 0.
A restricted interval changes that reasoning. Check whether x = h lies inside it, and evaluate the included endpoints. Use the smallest and largest values actually reached.
For f(x) = (x − 1)² − 4:
Use the domain selector in the graph section to compare these three cases. A closed endpoint includes its value; an open endpoint may exclude it unless another allowed input produces the same output.
g(x) = −3x² − 12x − 15
The coefficient of x² is negative.
g(x) = −3(x + 2)² − 3
Complete the square.
Maximum: −3 at x = −2
For all real x, the range is y ≤ −3.
No real x-intercepts
The whole graph is below the x-axis. Its y-intercept is (0, −15).
04 / Find the equation
A turning point gives the form a(x − h)² + k. A pair of roots r and s gives a(x − r)(x − s). One other point can then determine a.
A graph has roots −2 and 4 and passes through (0, 16).
y = a(x + 2)(x − 4)
16 = a(2)(−4), so a = −2
y = −2x² + 4x + 16
Its axis is halfway between the roots: x = 1. Substituting gives the turning point (1, 18).
The roots alone do not fix the scale a. Many parabolas can share the same two crossings.
Vertex (−1, 6); point (1, −2)
Start with the turning-point form.
y = a(x + 1)² + 6
Substitute the second point.
−2 = 4a + 6 ⇒ a = −2
The curve opens downwards.
y = −2x² − 4x + 4
Expand to read the coefficients. Check y(−1) = 6 and y(1) = −2.
05 / Intersections
At an intersection, f(x) and g(x) have the same value at the same x. Solve f(x) = g(x), then substitute each x into either function to recover y.
Do not stop at the input values if the question asks for coordinates. Check with the other function.
f(x) = x³ − 4x, g(x) = 2x² − x
f(x) = g(x) ⇒ x³ − 2x² − 3x = 0
x(x − 3)(x + 1) = 0
x = −1, 0, 3
The intersection coordinates are (−1, 3), (0, 0) and (3, 15). Factor out x; do not divide by it.
f(x) = x² − 2x − 3
g(x) = 2x + 2
Set the expressions equal.
x² − 4x − 5 = 0
(x − 5)(x + 1) = 0
The two inputs are x = 5 and x = −1.
g(5) = 12; g(−1) = 0
Recover the output for each input.
Intersections: (5, 12), (−1, 0)
Both pairs satisfy f and g.
06 / Your turn
Sketch before checking the worked solution. Label coordinates, not just the numbers on the axes.
For f(x) = 2x² − x + 1, find f(−2) and f(t + 1). Give the domain of g(x) = √(5 − x).
Bracket the entire substituted input. A real square root needs a non-negative radicand.
f(−2) = 8 + 2 + 1 = 11
f(t + 1) = 2(t + 1)² − (t + 1) + 1
= 2t² + 3t + 2
For g, 5 − x ≥ 0, so x ≤ 5.
Find the intercepts, turning point and symmetry line of y = −x² + 6x − 5.
Factorise for the roots and complete the square for the vertex.
y = −(x − 1)(x − 5)
y = 4 − (x − 3)²
Intercepts: (1, 0), (5, 0), (0, −5). Maximum at (3, 4); axis x = 3. Draw a smooth downward parabola through these points.
Find the range of f(x) = (x − 3)² + 2 for −1 ≤ x ≤ 2.
The turning point is outside the domain. Compare the endpoint values and the direction of the curve.
f(−1) = 18; f(2) = 3
The curve decreases throughout this interval because its inputs are to the left of x = 3. Range: 3 ≤ f(x) ≤ 18.
A parabola has minimum (2, −7) and passes through (0, 1). Find its equation.
Write y = a(x − 2)² − 7, then use (0, 1).
1 = 4a − 7 ⇒ a = 2
y = 2(x − 2)² − 7
= 2x² − 8x + 1
Find the intersections of y = x² + x − 6 and y = 3x + 2.
Equate the right-hand sides and solve for x. Then find both y values.
x² − 2x − 8 = 0
(x − 4)(x + 2) = 0
x = 4 or x = −2
The coordinates are (4, 14) and (−2, −4).
Show that 2x² − 8x + 11 is positive for every real x.
Complete the square and use the fact that a square is non-negative.
2x² − 8x + 11 = 2(x − 2)² + 3 ≥ 3
Its minimum is 3, which is strictly positive. Testing a few inputs would not prove the claim for all real x.
07 / Recap
Section 1 of 7 · Functions