2 Functions

Use gradient and intercepts to write a line.

A line can be written y=mx+c; m measures change in y per unit x and c is the y-intercept.

Worked example
Through (1,3) and (3,7), m=2 and c=1, so y=2x+1.

Worked example
Are lines with equal gradients parallel? yes, unless they are the same line.

Common boundary
Perpendicular gradients are negative reciprocals, not simply opposite signs.

Treat a function as a rule with a defined input set.

A function assigns exactly one output to each allowed input; domain is inputs, range is resulting outputs, and the graph shows the relation.

Worked example
For f(x)=√x, the real domain is x≥0.

Worked example
Can one input have two outputs? not in a function.

Common boundary
A graph can fail the vertical-line test even if it looks like a curve.

Inverse example: if $f(x)=2x-3$, write $y=2x-3$, swap $x$ and $y$, and solve to get $f^{-1}(x)=(x+3)/2$. The graphs of $f$ and $f^{-1}$ reflect in $y=x$, and the domain of $f^{-1}$ is the range of $f$. An inverse function exists only after the original function is one-to-one on its stated domain.

Sketch a graph from structure and context.

Use intercepts, turning points, asymptotes, scale and context to sketch the shape before relying on a calculator.

Worked example
A revenue graph crossing the x-axis at 0 and 100 has break-even intercepts; label what each axis means.

Worked example
What makes a sketch usable? labelled axes and key features, not artistic smoothness.

Common boundary
A calculator trace without scale or labels is not a mathematical interpretation.

Sums and differences are graphed pointwise: if $f(x)=x^2$ and $g(x)=2x$, then $(f+g)(x)=x^2+2x$ and $(f-g)(x)=x^2-2x$. Generate or sketch each new function, then label axes, intercepts, extrema and any context-specific units. Do not add the visual heights without first checking both functions share the same input domain.

Read key features as evidence about a function.

Intercepts, roots, extrema, symmetry, vertices and asymptotes answer different questions about a graph.

Worked example
For y=(x−2)²−9, the vertex is (2,−9) and roots are x=−1,5.

Worked example
Which feature gives break-even inputs? x-intercepts/roots.

Common boundary
A minimum point is not automatically an x-intercept.

Compose functions in the stated order.

(f∘g)(x)=f(g(x)); composition feeds the output of one rule into the next, so order matters.

Worked example
If f(x)=2x and g(x)=x+3, f∘g=2x+6 but g∘f=2x+3.

Worked example
Why can the two compositions differ? the inner function is applied first.

Common boundary
Composition is not ordinary multiplication.

Finding and checking an inverse: for $f(x)=3x-5$, solve $y=3x-5$ for $x$, giving $f^{-1}(x)=(x+5)/3$. Then $(f\circ f^{-1})(x)=3[(x+5)/3]-5=x$ and $(f^{-1}\circ f)(x)=[(3x-5)+5]/3=x$. State any domain restriction needed to make a non-one-to-one function invertible.

Move between quadratic forms to expose meaning.

Standard, factorized and vertex forms reveal different features: intercept, roots, and vertex/axis.

Worked example
x²−6x+5=(x−1)(x−5)=(x−3)²−4.

Worked example
Which form shows roots immediately? factorized form.

Common boundary
Do not read the vertex from the constant term in standard form.

Solve quadratic equations and inequalities with sign control.

Use factorization, completing the square or the formula for equations; for inequalities, test intervals or use the parabola’s sign.

Worked example
(x−2)(x+1)≥0 gives x≤−1 or x≥2.

Worked example
Why are there two intervals? the product is positive outside the roots for an upward parabola.

Common boundary
Solving the equation alone does not solve the inequality.

Quadratic formula and discriminant: for $ax^2+bx+c=0$, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ and $\Delta=b^2-4ac$. If $\Delta>0$ there are two distinct real roots, $\Delta=0$ gives one repeated real root, and $\Delta<0$ gives no real roots. For $2x^2-3x-2=0$, $\Delta=25$, so $x=(3\pm5)/4$, giving $x=2$ or $x=-1/2$.

Use asymptotes and intercepts to sketch rational functions.

For a rational function, excluded denominator zeros create vertical asymptotes; end behaviour gives horizontal or oblique asymptotes.

Worked example
f(x)=1/(x−2)+3 has vertical asymptote x=2 and horizontal asymptote y=3.

Worked example
Can the graph cross a vertical asymptote? no, the function is undefined there.

Common boundary
An asymptote is not always a line the graph never approaches closely.

Linear-over-linear example: for $f(x)=\frac{2x-3}{x+4}$, the vertical asymptote is $x=-4$ and the horizontal asymptote is $y=2$ (the ratio of leading coefficients). The x-intercept is $x=3/2$ and the y-intercept is $f(0)=-3/4$. Plot these features before sketching the branches; $x=-4$ is excluded from the domain.

Use inverse structure to connect exponential and logarithmic graphs.

Exponential and logarithmic functions are inverses, so their graphs reflect in y=x and have linked domain/range restrictions.

Worked example
y=2ˣ has range y>0; y=log₂x has domain x>0.

Worked example
What does the reflection do? swaps x and y, turning one inverse graph into the other.

Common boundary
The logarithm is not defined for non-positive real inputs.

Choose graphical or analytic equation solving deliberately.

An equation can be solved by algebra when structure permits, or by graph intersection/technology when an exact analytic route is unsuitable.

Worked example
x²=5 gives ±√5 analytically; eˣ=3 can be reported as ln3 or approximated.

Worked example
What does an intersection represent? equal y-values, hence a solution to f(x)=g(x).

Common boundary
A calculator decimal is not automatically the exact solution.

Apply graph transformations in the correct order.

Translations change position, reflections change orientation and stretches change scale; the order matters when transformations are composed.

Worked example
y=2f(x−3) shifts right 3 then stretches vertically by 2.

Worked example
What does f(x−3) do? move the graph right 3, not left.

Common boundary
Inside changes act horizontally with reversed sign; outside changes act vertically.

Use factor and remainder theorems to test polynomial roots.

For a polynomial p(x), (x−a) is a factor exactly when p(a)=0; the remainder on division by (x−a) is p(a). The coefficients and roots are linked by Vieta relationships.

Corrected worked example

For $p(x)=x^3-4x^2+x+6$, $p(2)=0$, so $x-2$ is a factor. Division gives $p(x)=(x-2)(x^2-2x-3)=(x-2)(x-3)(x+1)$. The roots are $2$, $3$ and $-1$; none is repeated.

For this cubic, the roots sum to $2+3-1=4=-(-4)/1$ and their product is $2\cdot3\cdot(-1)=-6=(-1)^3(6/1)$. Use the factor test at the candidate value, then verify the full factorization and Vieta relations.

A zero of p is a number a with p(a)=0; x=0 is not the same statement as the factor x.

For $a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0=0$, the sum of all roots (with multiplicity) is $-a_{n-1}/a_n$ and their product is $(-1)^na_0/a_n$. A repeated root requires a repeated factor; one successful factor test alone does not prove multiplicity.

Read asymptotes from a rational function’s leading structure.

Vertical asymptotes occur at non-cancelled denominator zeros; horizontal or oblique behaviour comes from comparing numerator and denominator degrees after simplification.

Worked example

For f(x)=(x²+1)/(x−2), x=2 is a vertical asymptote. Polynomial division gives f(x)=x+2+5/(x−2), so y=x+2 is the oblique asymptote.

A cancelled factor creates a hole, not a vertical asymptote; always simplify and record excluded domain values.

An asymptote describes limiting behaviour, not a value the function reaches at the asymptote.

Complete the graph-feature check for $f(x)=\frac{x^2+1}{x-2}=x+2+\frac5{x-2}$. Besides vertical asymptote $x=2$ and oblique asymptote $y=x+2$, the y-intercept is $f(0)=-1/2$. There are no real x-intercepts because $x^2+1=0$ has no real solution. Record all intercepts, asymptotes, holes and excluded inputs before sketching.

Use symmetry and domain restrictions to analyse inverses.

An even function satisfies f(−x)=f(x), an odd function satisfies f(−x)=−f(x). An inverse exists as a function only after the original is one-to-one on its chosen domain.

Worked example

f(x)=x² is even but not one-to-one on ℝ. Restricting to x≥0 gives f⁻¹(x)=√x; the graph reflects across y=x.

Check the domain before finding an inverse; the same formula can produce different inverse branches.

Symmetry does not imply invertibility: even functions usually map two inputs to one output.

Periodic and self-inverse examples: $\cos x$ is even and $2\pi$-periodic, while $\sin x$ is odd and $2\pi$-periodic. The function $f(x)=1/x$ on $x\ne0$ is odd and self-inverse because $f(f(x))=1/(1/x)=x$. Self-inverse means $f^{-1}=f$; it does not mean every input is fixed by $f$.

Solve an inequality by comparing graphs or sign intervals.

To solve g(x)≥f(x), find where h(x)=g(x)−f(x) is non-negative. Intersections split the number line into intervals whose signs must be checked.

Worked example

For x²≥2x, x(x−2)≥0, so x≤0 or x≥2. The graph gives the same result because the parabola lies above the line outside the intersections.

Include equality at roots for ≥ or ≤, and exclude points where an expression is undefined.

The intersection points are boundaries; they are not automatically the only solutions.

Apply modulus and reciprocal transformations in the correct order.

For y=|f(x)|, negative parts reflect above the x-axis; y=f(|x|) mirrors the right-hand graph into x<0; y=1/f(x) keeps zeros as excluded inputs and swaps large/small values.

Worked example

If f(x)=x−1, then |f(x)| has a corner at x=1, while f(|x|)=|x|−1 has a V-shaped graph with a corner at x=0.

Transform the graph in stages and preserve domain restrictions; the two modulus forms are not interchangeable.

Absolute value outside changes outputs; absolute value inside changes which inputs are used.

Further transformations: $y=[f(x)]^2$ makes outputs non-negative and retains the zeros of $f$; $y=f(ax+b)$ applies horizontal scaling and translation through the input. For example, $f(2x-4)=f(2(x-2))$ is horizontally compressed by factor $1/2$ and shifted right $2$. Modulus example: $|2x-3|\le5$ is equivalent to $-5\le2x-3\le5$, giving $-1\le x\le4$.