Functions and Their Graphs

Study Sheet

Functions and Their Graphs

Notation, domain & range, graph analysis, parent functions, transformations, combinations, inverses, and piecewise functions

Function Notation, Domain & Range, Evaluating

Concept
What is a function?

A function ff assigns to each input xx (from the domain) exactly one output y=f(x)y=f(x). The set of all outputs is the range.

  • Evaluating: substitute the input for every xx. In f(x)=x2−3xf(x)=x^2-3x, we get f(a+1)=(a+1)2−3(a+1)f(a+1)=(a+1)^2-3(a+1).
  • Domain of a polynomial: all reals, (−∞,∞)(-\infty,\infty).
  • Radical g(x)\sqrt{g(x)}: require g(x)≥0g(x)\ge 0.
  • Rational p(x)q(x)\dfrac{p(x)}{q(x)}: require q(x)≠0q(x)\ne 0.

Precalculus treats functions as objects: their domain and range, where they increase, their symmetry (even, odd), their transformations, their compositions and inverses, and their average rate of change on an interval. Piecewise functions and the parent-function library (linear, quadratic, cubic, root, reciprocal, absolute value) are the vocabulary for everything later.

Example
Finding a domain

Find the domain of f(x)=x+2x−3f(x)=\dfrac{\sqrt{x+2}}{x-3}.

Need x+2≥0⇒x≥−2x+2\ge 0 \Rightarrow x\ge -2, and x−3≠0⇒x≠3x-3\ne 0 \Rightarrow x\ne 3.

Domain: [−2,3)∪(3,∞)[-2,3)\cup(3,\infty).

Side note
Where it goes wrong

Composing in the wrong order. (f∘g)(x)=f(g(x))(f\circ g)(x) = f(g(x)) applies gg FIRST; f∘gf\circ g and g∘fg\circ f are usually different functions.

Example
Evaluating

For f(x)=x2−3xf(x)=x^2-3x, compute f(−2)f(-2) and f(a+1)f(a+1).

f(−2)=(−2)2−3(−2)=4+6=10f(-2)=(-2)^2-3(-2)=4+6=10.

f(a+1)=(a+1)2−3(a+1)=a2+2a+1−3a−3=a2−a−2f(a+1)=(a+1)^2-3(a+1)=a^2+2a+1-3a-3=a^2-a-2.

Tip

Tip: A domain restriction from a radical uses ≥\ge; a restriction from a denominator uses ≠\ne. Always combine both kinds of restriction when they appear together.

Try it
Quick check

Find the domain of g(x)=6−2xx+1g(x)=\dfrac{\sqrt{6-2x}}{x+1}.

Show the answer

Answer: (−∞,−1)∪(−1,3](-\infty,-1)\cup(-1,3]. The radical needs 6−2x≥06-2x\ge0, so x≤3x\le3; the denominator needs x≠−1x\ne-1. Both restrictions apply at once.

The Difference Quotient

Concept
Difference quotient

The difference quotient of ff is

f(x+h)−f(x)h,h≠0.\frac{f(x+h)-f(x)}{h},\qquad h\ne 0.

It measures the average rate of change of ff over [x,x+h][x,x+h] and is the foundation of the derivative. Simplify until the hh in the denominator cancels.

The two points (x,f(x))(x,f(x)) and (x+h,f(x+h))(x+h,f(x+h)) on the graph are hh apart horizontally, so the difference quotient is the slope of the line through them. Simplifying until hh leaves the denominator is what lets us later ask what that slope approaches as the two points merge, which is the derivative.

Example
Difference quotient of a quadratic

Let f(x)=x2+1f(x)=x^2+1. Then

f(x+h)=(x+h)2+1=x2+2xh+h2+1,f(x+h)=(x+h)^2+1=x^2+2xh+h^2+1,
f(x+h)−f(x)h=(x2+2xh+h2+1)−(x2+1)h=2xh+h2h=2x+h.\frac{f(x+h)-f(x)}{h}=\frac{(x^2+2xh+h^2+1)-(x^2+1)}{h}=\frac{2xh+h^2}{h}=2x+h.
Tip

Tip: The f(x+h)−f(x)f(x+h)-f(x) numerator should always lose its constant term and its lone x2x^2 term, leaving only factors of hh. If no hh cancels, recheck the algebra.

Try it
Quick check

Find and simplify the difference quotient of f(x)=3x2−xf(x)=3x^2-x.

Show the answer

Answer: 6x+3h−16x+3h-1. f(x+h)=3x2+6xh+3h2−x−hf(x+h)=3x^2+6xh+3h^2-x-h. Subtracting f(x)f(x) leaves 6xh+3h2−h6xh+3h^2-h, and dividing by hh gives 6x+3h−16x+3h-1.

Analyzing Graphs

Concept
Reading a graph
  • xx-intercepts (zeros): where f(x)=0f(x)=0; yy-intercept: the value f(0)f(0).
  • Even (symmetric about yy-axis): f(−x)=f(x)f(-x)=f(x).   Odd (symmetric about origin): f(−x)=−f(x)f(-x)=-f(x).
  • Increasing / decreasing / constant over an interval as xx moves left to right.
  • Relative maximum / minimum: a peak or valley of the graph.
Example
Analyzing f(x)=x3−3xf(x)=x^3-3x
xy−2−11−3−2−112(−1, 2)(1, −2)
Zeros at 0,±30,\pm\sqrt3; turning points (−1,2)(-1,2) and (1,−2)(1,-2); symmetric about the origin.
  • Zeros: x3−3x=x(x2−3)=0⇒x=0, ±3x^3-3x=x(x^2-3)=0 \Rightarrow x=0,\ \pm\sqrt{3}.
  • Symmetry: f(−x)=−x3+3x=−f(x)f(-x)=-x^3+3x=-f(x), so ff is odd.
  • Relative max (−1,2)(-1,2); relative min (1,−2)(1,-2).
  • Increasing on (−∞,−1)(-\infty,-1) and (1,∞)(1,\infty); decreasing on (−1,1)(-1,1).
Tip

Tip: To test symmetry algebraically, substitute −x-x. If you get the same expression it is even; if you get the opposite of the original it is odd; otherwise neither.

Try it
Quick check

For f(x)=x4−2x2f(x)=x^4-2x^2: is ff even, odd, or neither? Find its zeros and its relative extrema (they are at x=0x=0 and x=±1x=\pm1).

Show the answer

Answer: even; zeros 00 (touching) and ±2\pm\sqrt2; relative max (0,0)(0,0), relative minima (±1,−1)(\pm1,-1). f(−x)=x4−2x2=f(x)f(-x)=x^4-2x^2=f(x). Factor: x2(x2−2)=0x^2(x^2-2)=0. At x=±1x=\pm1, f=1−2=−1f=1-2=-1.

The Library of Parent Functions

Concept
Eight parents to memorize

Constant f(x)=cf(x)=c; Identity f(x)=xf(x)=x; Absolute value f(x)=∣x∣f(x)=|x|; Square root f(x)=xf(x)=\sqrt{x}; Quadratic f(x)=x2f(x)=x^2; Cubic f(x)=x3f(x)=x^3; Reciprocal f(x)=1xf(x)=\dfrac{1}{x}; Greatest integer (step) f(x)=⌊x⌋f(x)=\lfloor x\rfloor.

The eight parent graphs, each on the window −2≤x≤2-2\le x\le 2.

Tip

Tip: x\sqrt{x} has domain [0,∞)[0,\infty) and 1x\tfrac{1}{x} excludes x=0x=0 (a vertical asymptote). All other parents above have domain (−∞,∞)(-\infty,\infty).

Try it
Quick check

Which of the eight parent functions are odd (symmetric about the origin)? Which are even?

Show the answer

Answer: odd: xx, x3x^3, 1x\tfrac1x. Even: ∣x∣|x|, x2x^2, and the constant function. Test with f(−x)f(-x): (−x)3=−x3(-x)^3=-x^3 and 1−x=−1x\tfrac{1}{-x}=-\tfrac1x flip sign; ∣−x∣=∣x∣|-x|=|x| and (−x)2=x2(-x)^2=x^2 do not. x\sqrt x and ⌊x⌋\lfloor x\rfloor are neither (−x\sqrt{-x} is not even defined for x>0x>0, and ⌊−0.5⌋=−1≠−⌊0.5⌋\lfloor-0.5\rfloor=-1\ne-\lfloor0.5\rfloor).

Transformations of Functions

Concept
Reading y=a f(bx−c)+dy=a\,f(bx-c)+d

Starting from y=f(x)y=f(x):

  • +d+d: shift up dd;   −d-d: shift down.
  • f(x−c)f(x-c): shift right cc;   f(x+c)f(x+c): shift left. (Horizontal moves are opposite the sign.)
  • a<0a<0: reflect over the xx-axis;   ∣a∣>1|a|>1: vertical stretch;   0<∣a∣<10<|a|<1: vertical shrink.
  • b<0b<0: reflect over the yy-axis;   ∣b∣>1|b|>1: horizontal shrink;   0<∣b∣<10<|b|<1: horizontal stretch.

Inside changes act on xx before the function sees it, so they are undone backwards: f(x−2)f(x-2) reaches the value f(0)f(0) only when x=2x=2, which moves the graph right, not left. Outside changes act on the output after the fact, so they move the graph the way they look: +3+3 raises every point by 33.

Example
A multi-step transformation
xy−3−2−1123−2−1123(1, 3)
The vertex travels from (0,0)(0,0) to (1,3)(1,3) and the V turns upside down.

Graph g(x)=−∣x−1∣+3g(x)=-|x-1|+3 from the parent f(x)=∣x∣f(x)=|x|: shift right 11, reflect over the xx-axis, then shift up 33. The vertex moves from (0,0)(0,0) to (1,3)(1,3) and the graph opens downward. In the figure, ff is dashed (pre-image) and gg is solid (image).

Tip

Tip: Apply transformations in order: horizontal shift/stretch (inside the function) first, then vertical stretch/reflection, then vertical shift. When b≠1b\ne 1, factor first: f(bx−c)=f ⁣(b(x−cb))f(bx-c)=f\!\big(b(x-\tfrac{c}{b})\big) so the horizontal shift is c/bc/b, not cc.

Try it
Quick check

Describe how g(x)=2x+3−1g(x)=2\sqrt{x+3}-1 is obtained from x\sqrt x, and give its starting point, domain and range.

Show the answer

Answer: shift left 33, stretch vertically by 22, shift down 11. Starting point (−3,−1)(-3,-1); domain [−3,∞)[-3,\infty); range [−1,∞)[-1,\infty). The parent's endpoint (0,0)(0,0) moves left 33 and down 11; the stretch does not move it because its height is 00.

Combinations and Composition of Functions

Concept
Arithmetic and composition

For functions ff and gg:

(f±g)(x)=f(x)±g(x),(fg)(x)=f(x) g(x),(fg)(x)=f(x)g(x), g(x)≠0.(f\pm g)(x)=f(x)\pm g(x),\quad (fg)(x)=f(x)\,g(x),\quad \left(\tfrac{f}{g}\right)(x)=\frac{f(x)}{g(x)},\ g(x)\ne 0.

The composition (f∘g)(x)=f(g(x))(f\circ g)(x)=f\big(g(x)\big) feeds the output of gg into ff.

Example
Composition and its domain

Let f(x)=xf(x)=\sqrt{x} and g(x)=x−4g(x)=x-4. Then

(f∘g)(x)=f(x−4)=x−4.(f\circ g)(x)=f(x-4)=\sqrt{x-4}.

The domain needs x−4≥0x-4\ge 0, so domain=[4,∞)\textbf{domain}=[4,\infty).   Also (g∘f)(x)=x−4(g\circ f)(x)=\sqrt{x}-4, with domain [0,∞)[0,\infty).

Tip

Tip: The domain of f∘gf\circ g is the set of xx in the domain of gg whose outputs g(x)g(x) lie in the domain of ff. Composition is not commutative: usually f∘g≠g∘ff\circ g\ne g\circ f.

Try it
Quick check

Let f(x)=x2+1f(x)=x^2+1 and g(x)=x−2g(x)=\sqrt{x-2}. Find (f∘g)(x)(f\circ g)(x) and its domain.

Show the answer

Answer: (f∘g)(x)=x−1(f\circ g)(x)=x-1 with domain [2,∞)[2,\infty). f(g(x))=(x−2)2+1=x−2+1=x−1f(g(x))=(\sqrt{x-2})^2+1=x-2+1=x-1. The simplified formula looks defined everywhere, but gg must be defined first, so x≥2x\ge2.

Inverse Functions

Concept
One-to-one and inverses

ff is one-to-one if each output comes from exactly one input; equivalently it passes the horizontal line test. Only one-to-one functions have an inverse f−1f^{-1}, which satisfies

f(f−1(x))=xandf−1(f(x))=x.f\big(f^{-1}(x)\big)=x \quad\text{and}\quad f^{-1}\big(f(x)\big)=x.

The graph of f−1f^{-1} is the reflection of the graph of ff over the line y=xy=x.

An inverse runs the function backwards, sending each output to the input it came from. That only works if no output came from two inputs, which is the horizontal-line test. Since running backwards swaps every point (a,b)(a,b) to (b,a)(b,a), the graph of f−1f^{-1} is the reflection in y=xy=x, and finding f−1f^{-1} algebraically is the same swap: exchange xx and yy, then solve.

Example
Finding and verifying an inverse
xy−6−4−224−6−4−224ff⁻¹y = x
f(x)=2x+6f(x)=2x+6 and f−1(x)=x−62f^{-1}(x)=\tfrac{x-6}{2} are mirror images in y=xy=x: (−3,0)(-3,0) on ff becomes (0,−3)(0,-3) on f−1f^{-1}.

Find the inverse of f(x)=2x+6f(x)=2x+6.   Write y=2x+6y=2x+6, swap: x=2y+6x=2y+6, solve: y=x−62y=\dfrac{x-6}{2}.

So f−1(x)=x−62f^{-1}(x)=\dfrac{x-6}{2}. Verify: f ⁣(f−1(x))=2 ⁣⋅ ⁣x−62+6=xf\!\big(f^{-1}(x)\big)=2\!\cdot\!\dfrac{x-6}{2}+6=x. (verified)

Tip

Tip: If a function fails the horizontal line test (like f(x)=x2f(x)=x^2), restrict its domain (e.g. x≥0x\ge 0) to make it one-to-one before inverting.

Try it
Quick check

Find the inverse of f(x)=x−1x+3f(x)=\dfrac{x-1}{x+3}.

Show the answer

Answer: f−1(x)=3x+11−xf^{-1}(x)=\dfrac{3x+1}{1-x}. Swap: x=y−1y+3x=\dfrac{y-1}{y+3}, so xy+3x=y−1xy+3x=y-1, then y(x−1)=−3x−1y(x-1)=-3x-1 and y=3x+11−xy=\dfrac{3x+1}{1-x}. Check: f(0)=−13f(0)=-\tfrac13 and f−1 ⁣(−13)=04/3=0f^{-1}\!\left(-\tfrac13\right)=\dfrac{0}{4/3}=0.

Piecewise Functions and Average Rate of Change

Concept
Piecewise functions
xy−2−11234−1123
Open dot (1,2)(1,2): excluded; closed dot (1,3)(1,3): f(1)=3f(1)=3 comes from the second rule.

A piecewise function uses different formulas on different parts of the domain. A closed dot ∙\bullet includes an endpoint; an open dot ∘\circ excludes it. Evaluate by choosing the rule whose condition the input satisfies.

f(x)={x+1,x<14−x,x≥1f(x)=\begin{cases} x+1, & x<1 \\[2pt] 4-x, & x\ge 1 \end{cases}

Here f(0)=1f(0)=1 and f(1)=3f(1)=3.

Concept
Average rate of change
xy1232468slope 4
The secant through (1,1)(1,1) and (3,9)(3,9) has slope 9−13−1=4\tfrac{9-1}{3-1}=4.

The average rate of change of ff from x=ax=a to x=bx=b is the slope of the secant line:

f(b)−f(a)b−a.\frac{f(b)-f(a)}{b-a}.

It is the constant rate that would carry the function from its starting value to its ending value over the same interval, just as an average speed is the constant speed that covers the same distance in the same time. Geometrically it is the slope of the secant line, and it says nothing about what happens between the two endpoints.

Example
Average rate of change

For f(x)=x2f(x)=x^2 from a=1a=1 to b=3b=3:

f(3)−f(1)3−1=9−12=4.\frac{f(3)-f(1)}{3-1}=\frac{9-1}{2}=4.
Tip

Tip: Average rate of change is the difference quotient with x=ax=a and h=b−ah=b-a; it is a secant slope, not the value of the function.

Try it
Quick check

Find the average rate of change of f(x)=x3f(x)=x^3 from x=−1x=-1 to x=2x=2.

Show the answer

Answer: 33. f(2)−f(−1)2−(−1)=8−(−1)3=93=3\dfrac{f(2)-f(-1)}{2-(-1)}=\dfrac{8-(-1)}{3}=\dfrac93=3.

Going Deeper: Advanced Function Ideas

Concept
Functional equations

A functional equation constrains an unknown function ff by an identity that must hold for all allowed xx. The key tactic is substitution: replace xx with a related expression (such as 1x\tfrac{1}{x}, −x-x, or 1−x1-x) to generate a second equation, then solve the resulting system for f(x)f(x) as if f(x)f(x) and the substituted value were two separate unknowns.

  • Common self-inverse swaps: x↦1xx\mapsto \tfrac{1}{x}, x↦−xx\mapsto -x, x↦1−xx\mapsto 1-x.
  • After substituting, treat f(x)f(x) and f(new input)f(\text{new input}) as variables and eliminate.
Example
Solving a functional equation

Find ff if f(x)+2f ⁣(1x)=3xf(x)+2f\!\left(\tfrac{1}{x}\right)=3x for all x≠0x\ne 0.

Replace xx with 1x\tfrac{1}{x} to get a companion equation, giving the system

f(x)+2f ⁣(1x)=3x,2f(x)+f ⁣(1x)=3x.\begin{aligned} f(x)+2f\!\left(\tfrac{1}{x}\right) &= 3x, \\[2pt] 2f(x)+f\!\left(\tfrac{1}{x}\right) &= \tfrac{3}{x}. \end{aligned}

Multiply the second by 22 and subtract the first: 3f(x)=6x−3x3f(x)=\dfrac{6}{x}-3x, so

f(x)=2x−x.f(x)=\frac{2}{x}-x.

Check: f(x)+2f ⁣(1x)=(2x−x)+2(2x−1x)=3x.f(x)+2f\!\left(\tfrac1x\right)=\left(\tfrac2x-x\right)+2\left(2x-\tfrac1x\right)=3x. (verified)

Concept
Iterated compositions and involutions

Composing ff with itself nn times is written

f[n]=f∘f∘⋯∘f⏟n copies,f[1]=f,  f[n+1]=f∘f[n].f^{[n]}=\underbrace{f\circ f\circ\cdots\circ f}_{n\ \text{copies}},\qquad f^{[1]}=f,\ \ f^{[n+1]}=f\circ f^{[n]}.

(The bracket distinguishes this from a power f(x)nf(x)^n or the inverse f−1f^{-1}.) A function is an involution if f∘f=idf\circ f=\mathrm{id}, i.e. f(f(x))=xf\big(f(x)\big)=x; then f−1=ff^{-1}=f and the iterates cycle: f[even]=idf^{[\text{even}]}=\mathrm{id} and f[odd]=ff^{[\text{odd}]}=f. Some maps have longer periods (e.g. f(x)=1−1xf(x)=1-\tfrac{1}{x} satisfies f[3]=idf^{[3]}=\mathrm{id}).

Example
An involution and its iterates

Let f(x)=xx−1f(x)=\dfrac{x}{x-1} with x≠1x\ne 1. Compute f∘ff\circ f:

f(f(x))=xx−1xx−1−1=xx−1x−(x−1)x−1=xx−11x−1=x.f\big(f(x)\big)=\frac{\frac{x}{x-1}}{\frac{x}{x-1}-1} =\frac{\frac{x}{x-1}}{\frac{x-(x-1)}{x-1}} =\frac{\frac{x}{x-1}}{\frac{1}{x-1}}=x.

So ff is an involution: f[2]=idf^{[2]}=\mathrm{id} and f−1=ff^{-1}=f. The iterates therefore alternate, and for instance f[100]=idf^{[100]}=\mathrm{id} while f[101]=ff^{[101]}=f.

Concept
Monotonicity, symmetry, and composed transformations

Three high-leverage ideas:

  • Monotonicity ⇒\Rightarrow invertibility. If ff is strictly increasing (or strictly decreasing) on its domain, it automatically passes the horizontal line test, so f−1f^{-1} exists there — no need to solve algebraically to know it is one-to-one.
  • Exploiting even/odd symmetry. An even function is fully determined by its values on [0,∞)[0,\infty); an odd function must have f(0)=0f(0)=0 (when 00 is in the domain). Products/quotients obey: even⋅\cdoteven == even, odd⋅\cdotodd == even, even⋅\cdotodd == odd.
  • Composing transformations into one mapping. Applying transformation T1T_1 then T2T_2 means the combined rule is T2∘T1T_2\circ T_1. Order matters: a shift-then-stretch gives a different single formula y=a f(bx−c)+dy=a\,f(bx-c)+d than stretch-then-shift.
Example
Domain of a nested composition

Find the domain of F(x)= 4−x F(x)=\sqrt{\,4-\sqrt{x}\,}, a composition F=f∘gF=f\circ g with f(u)=uf(u)=\sqrt{u} and g(x)=4−xg(x)=4-\sqrt{x}.

Work from the inside out, collecting every restriction:

  • Inner radical: x≥0x\ge 0.
  • Outer radical: 4−x≥0⇒x≤4⇒x≤164-\sqrt{x}\ge 0 \Rightarrow \sqrt{x}\le 4 \Rightarrow x\le 16.

Both must hold, so domain=[0,16]\textbf{domain}=[0,16].

Example
A parameter that makes a piecewise function continuous

For which cc is

f(x)={cx+1,x≤2x2−c,x>2f(x)=\begin{cases} cx+1, & x\le 2 \\[2pt] x^2-c, & x>2 \end{cases}

continuous at the split point x=2x=2?

The two branches must agree in the limit at x=2x=2:

c(2)+1⏟left value=(2)2−c⏟right limit ⟹ 2c+1=4−c ⟹ 3c=3 ⟹ c=1.\underbrace{c(2)+1}_{\text{left value}}=\underbrace{(2)^2-c}_{\text{right limit}} \ \Longrightarrow\ 2c+1=4-c\ \Longrightarrow\ 3c=3\ \Longrightarrow\ c=1.

With c=1c=1 both pieces give 33 at x=2x=2, so the graph joins without a jump.

Example
Difference quotient of a harder function

Compute the difference quotient of f(x)=xf(x)=\sqrt{x} using the conjugate:

x+h−xh=x+h−xh⋅x+h+xx+h+x=(x+h)−xh(x+h+x)=1x+h+x.\frac{\sqrt{x+h}-\sqrt{x}}{h} =\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} =\frac{(x+h)-x}{h\big(\sqrt{x+h}+\sqrt{x}\big)} =\frac{1}{\sqrt{x+h}+\sqrt{x}}.

The hh cancels once the radicals are rationalized. (For f(x)=1xf(x)=\tfrac{1}{x} the same idea gives f(x+h)−f(x)h=−1x(x+h)\dfrac{f(x+h)-f(x)}{h}=\dfrac{-1}{x(x+h)}.)

Tip

Tip: For nested radicals and rational compositions, never simplify before finding the domain — restrictions from an inner piece survive even if they cancel algebraically. And for iterated maps, always confirm a suspected involution or period by computing f∘ff\circ f (and f∘f∘ff\circ f\circ f), not by eyeballing the formula.

Why It Works

An even function satisfies f(−x)=f(x)f(-x) = f(x), so its graph is its own mirror image across the yy-axis; an odd function satisfies f(−x)=−f(x)f(-x) = -f(x), so rotating its graph 180∘180^\circ about the origin leaves it unchanged. Polynomials with only even powers are even and with only odd powers are odd — the names come from there. The average rate of change f(b)−f(a)b−a\tfrac{f(b) - f(a)}{b - a} is the slope of the secant line, and watching it as b→ab \to a is the first step toward the derivative.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
Function transformations
y=a f(b(x−h))+ky = a\,f\big(b(x-h)\big)+k

What it means. hh shifts right, kk shifts up, aa stretches vertically (flipping if negative), bb compresses horizontally.

Example. y=(x−2)2+3y=(x-2)^2+3 shifts y=x2y=x^2 right 22 and up 33.

Why it works. Inside changes act on the input before ff does, so they run backwards: replacing xx by x−hx-h means the graph reaches a given output hh units later, i.e. shifted right. Outside changes act after ff, so they behave as written.

Tip. Inside the function: opposite of what it looks like. Outside: exactly what it looks like.

Tip
Inverse functions
f(f−1(x))=xandf−1(f(x))=xf\big(f^{-1}(x)\big)=x \quad\text{and}\quad f^{-1}\big(f(x)\big)=x

What it means. The inverse undoes the function; their graphs are mirror images across y=xy=x.

Example. f(x)=2x+1f(x)=2x+1 has inverse f−1(x)=x−12f^{-1}(x)=\tfrac{x-1}{2}, and f(f−1(x))=xf(f^{-1}(x))=x.

Why it works. Swapping xx and yy in y=f(x)y=f(x) and solving for yy produces the rule that reverses each input–output pair. Reflecting across y=xy=x is exactly what swapping coordinates does geometrically.

Tip. Only one-to-one functions have inverses. If a horizontal line hits the graph twice, restrict the domain first.