Porisma

Graphing techniques

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By the end, you can

  • Sketch rational curves using structural features
  • Find intercepts, turning points and asymptotes
  • Explain why a calculator window can hide features

about 40 minutes for the lesson · practise at your own pace.

The central idea

A sketch is a description of a function, not a screenshot. Find the domain, intercepts and symmetry, then explain its behaviour near excluded inputs and far from the origin. Polynomial division exposes horizontal or oblique asymptotes of rational functions.

Calculator entries for basic graphs

Enter each rule in Y=, use X,T,θ,n for the variable and GRAPH to draw it. The source calculator table gives:

FunctionEntry
x2x^2variable, square
x3x^3variable, power, 3
x\sqrt{x}2nd, x2x^2 (square-root template), variable
1/x1/x1, divide, variable
1/x21/x^21, divide, variable, square
x3\sqrt[3]{x}MATH, 4 (cube-root template), variable
exe^x2nd, LN (exponential template), variable
e−xe^{-x}exponential template, negative key, variable
ln⁡x\ln xLN, variable

The reciprocal entries also use ALPHA, Y= as a fraction-template alternative. To zoom into the cubic at (0,0)(0,0), position the zoom centre there and use ZOOM, 2, ENTER. Another calculator model may label these operations differently. Check the expression, domain and window.

Calculator mode and fraction entries

MODE: select FUNCTION with arrows and ENTER

  • MATHPRINT CLASSIC
  • NORMAL SCI ENG
  • FLOAT 0 1 2 3 4 5 6 7 8 9
  • RADIAN DEGREE
  • [FUNCTION] PARAMETRIC POLAR SEQ
  • THICK DOT-THICK THIN DOT-THIN
  • SEQUENTIAL SIMUL
  • REAL a+bi re^(θi)
  • FULL HORIZONTAL GRAPH-TABLE
  • FRACTION TYPE: n/d Un/d
  • ANSWERS: AUTO DEC FRAC-APPROX
  • GOTO 2ND FORMAT GRAPH: NO YES
  • STAT DIAGNOSTICS: OFF ON
  • STAT WIZARDS: ON OFF

The source highlights FUNCTION; its complete screen is in the evidence.

Fraction alternative: ALPHA → Y=

  • FRAC menu: 1:n/d 2:Un/d
  • 3: n/d ↔ Un/d 4: F ↔ D
  • Select 1:n/d; numerator 1, denominator X gives Y₁ = 1/X.
  • Use denominator X² for Y₁ = 1/X².

Teacher guidance: use TI-SmartView to introduce graphing settings.

Menu reconstruction retains source options; the date/clock is not a setting task.

MODE menu with FUNCTION, numeric/angle/format options and FRAC menu n/d, Un/d, conversions. Teacher recommends demonstrating settings with TI-SmartView.
Open original diagram (new tab)

Basic curves: square, cube and square root

y = x²

Y₁ = X²

Y= → X,T,θ,n → x² → GRAPH

Explore an inputx = 1
y = 1

Domain: All real inputs.

Parabola; vertex (0, 0).

y = x³

Y₁ = X^3

Y= → X,T,θ,n → ^ → 3 → GRAPH

Explore an inputx = 1
y = 1

Domain: All real inputs.

Near (0,0): ZOOM → 2 → ENTER (Zoom In).

y = √x

Y₁ = √(X)

Y= → 2nd → x² → X,T,θ,n → GRAPH

Explore an inputx = 1
y = 1

Domain: x ≥ 0.

Real inputs x ≥ 0; the curve begins at the origin.

Input/result reconstructions; viewing windows chosen for clear mathematical shape.

Calculator entry sequences and reconstructed x squared, x cubed and square-root curves. Cubic zoom at the origin; square root has domain x nonnegative.
Open original diagram (new tab)

Basic curves: reciprocals and cube root

y = 1/x

Y₁ = 1/X

Y= → 1 → ÷ → X,T,θ,n → GRAPH

Explore an inputx = 1
y = 1

Domain: x ≠ 0.

Two branches. Alternative entry: ALPHA → Y= (fraction template).

y = 1/x²

Y₁ = 1/X²

Y= → 1 → ÷ → X,T,θ,n → x² → GRAPH

Explore an inputx = 1
y = 1

Domain: x ≠ 0.

Both branches positive; the input 0 is excluded.

y = ∛x

Y₁ = ∛(X)

Y= → MATH → 4 → X,T,θ,n → GRAPH

Explore an inputx = 1
y = 1

Domain: All real inputs.

Real cube root: negative inputs give negative outputs.

Input/result reconstructions; viewing windows chosen for clear mathematical shape.

Calculator entry sequences for 1/x, 1/x squared and real cube root. Reciprocal branches split at zero; cube root includes negative inputs.
Open original diagram (new tab)

Basic curves: exponential and logarithm

y = eˣ

Y₁ = e^(X)

Y= → 2nd → LN → X,T,θ,n → GRAPH

Explore an inputx = 0
y = 1

Domain: All real inputs.

Increasing; y-intercept 1 and left-end asymptote y = 0.

y = e⁻ˣ

Y₁ = e^(−X)

Y= → 2nd → LN → (−) → X,T,θ,n → GRAPH

Explore an inputx = 0
y = 1

Domain: All real inputs.

Use the negation key (−); decreasing, with y-intercept 1.

y = ln x

Y₁ = ln(X)

Y= → LN → X,T,θ,n → GRAPH

Explore an inputx = 1
y = 0

Domain: x > 0.

Real inputs x > 0; x-intercept 1; vertical asymptote x = 0.

Input/result reconstructions; viewing windows chosen for clear mathematical shape.

Calculator entries for increasing e to x, decreasing e to negative x, and logarithm with positive domain and vertical asymptote x=0.
Open original diagram (new tab)

Read a logarithm beyond the sampled screen

Explore ln x

Change the view to see what coarse sampling misses.

Viewing window
Positive input x1
ln(1) = 0

Domain: x > 0. The curve passes through (1, 0).

Calculator notes

y = ln x: a finite sampled screen is not the whole curve

6:ZStandard: coarse positive-input samples in a −10..10 window

Analytical shape: x > 0, y → −∞ as x → 0⁺; zero at x = 1.

Use zoom options to reveal features, then justify them analytically.

Source ZOOM menu (visible rows):

  • 2:Zoom In 3:Zoom Out 4:ZDecimal
  • 5:ZSquare 6:ZStandard 7:ZTrig
  • 8:ZInteger 9:ZoomStat 0:ZoomFit

Cartesian diagrams show behaviour and relationships between variables.

Use conventions such as dotted asymptotes to identify the important features.

Sampling illustration, not an emulation of a specific calculator display.

Coarse logarithm samples compared with its analytical positive branch, plus source zoom-menu entries. Windows independently chosen; original calculator excerpts retained in review evidence.
Open original diagram (new tab)

Reflection axes: five basic curves

Test reflection symmetry

Choose a curve and a mirror line, then move point P.

Curve
Mirror line
Point input x1.5
P = (1.5, 2.25)P′ = (-1.5, 2.25)The reflected point lies on the curve.

A matching point is not a proof of symmetry.

Why this works

The y-axis, x = 0, is a reflection axis.

A matching point alone does not prove reflection symmetry. Compare the whole curve; the overlay is clipped to the viewing window. Solid and dashed lines identify the original and reflected curves.

Five source rows: upward parabola y=x² symmetric in x=0; sideways parabola x=y² in y=0; absolute-value V in x=0; reciprocal branches in y=x and y=−x; cosine with amplitudes ±1 and displayed axis x=0. Full cosine also has axes x=kπ.
Open original diagram (new tab)

Reflection axes: circle and cubic

Circle and cubic symmetry

Choose a curve and a mirror line, then move point P.

Curve
Mirror line
Line angle through origin0°
Point angle45°
P = (3.53553, 3.53553)P′ = (3.53553, -3.53553)The reflected point lies on the curve.

A matching point is not a proof of symmetry.

Why this works

Every straight line through the origin is a reflection axis. Equal axis scales keep this circle round.

A matching point alone does not prove reflection symmetry. Compare the whole curve; the overlay is clipped to the viewing window. Solid and dashed lines identify the original and reflected curves.

Circle x²+y²=25 with intercepts ±5 is symmetric in every line through the origin; horizontal and vertical examples shown. Cubic y=x³+x² has no reflection axis; intercept −1 and origin, with corrected exact maximum (−2/3,4/27). Printed −0.677 is explicitly recorded as a source slip.
Open original diagram (new tab)

From a line to absolute value

From a line to absolute value

Compare the original graph with its absolute value. The input stays the same; only negative outputs are reflected upwards.

Teal: Original graph. No comparison overlay.

Start: Original graph

y=xy=x

First identify the zeros and the parts above and below the x-axis.

Ideas to try
  1. Choose Step 1, then compare with Start. Which part of the graph changes?
  2. Open Point coordinates to check that the zeros stay fixed. Hide the comparison to inspect the final graph.
Source line y=x and reflected V y=|x| over x−3..3; negative original segment dashed; final panel labels y=−x for x<0 and y=x for x≥0.
Open original diagram (new tab)

Reflect only the negative quadratic section

Reflect only the negative quadratic section

Compare the original graph with its absolute value. The input stays the same; only negative outputs are reflected upwards.

Teal: Original graph. No comparison overlay.

Start: Original graph

y=x2−3x+2y=x^2-3x+2

First identify the zeros and the parts above and below the x-axis.

Ideas to try
  1. Choose Step 1, then compare with Start. Which part of the graph changes?
  2. Open Point coordinates to check that the zeros stay fixed. Hide the comparison to inspect the final graph.
Source quadratic x²−3x+2 with roots 1,2 and y-intercept 2; absolute-value curve reflects only its negative section between 1 and 2, showing the original arc dashed.
Open original diagram (new tab)

Explore

Compare the original rules y=2+c/(x−1)y=2+c/(x-1) for c=5c=5, c=−5c=-5 and c=0c=0, retaining x≠1x\ne1 in every case. Before graphing, predict the side of y=2y=2 on which each branch lies, the behaviour near x=1x=1 and the intercepts. Use a wide view and a close view near x=1x=1, then explain which features your screen supports and which require the algebra. For c=0c=0, decide why an excluded point remains even though there is no unbounded branch.

Related interactive example

Explore this related example, then return to the prediction task above.

Two rising branches

y = x − 1 − 2/(x − 2), with b = −3a and a = 1.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • C
  • Oblique asymptote y = a(x − 1)
  • Axis intercepts
Ideas to try
Why does the left branch never turn around?
Open the full interactive problem →

Worked example

For y=(2x+3)/(x−1)=2+5/(x−1)y=(2x+3)/(x-1)=2+5/(x-1), the domain excludes 1. The vertical asymptote is x=1x=1, the horizontal asymptote is y=2y=2, and intercepts are (−3/2,0)(-3/2,0) and (0,−3)(0,-3). The two branches approach the asymptotes with the sign of 5/(x−1)5/(x-1).

A graphing window does not prove that no other branch or turning point exists.

Check your understanding

A student sketches f(x)=(x2−1)/(x−2)f(x)=(x^2-1)/(x-2) using only −2≤x≤3-2\le x\le3 and claims that the curve has one turning point. Establish the original domain, both axial intercepts and the asymptotes. Use the derivative to assess the turning-point claim and explain what a more informative sketch must include.

Choose a practice question

The source’s PDF32 study pointers are retained below, including their broad classifications and printed subpart references. Follow the full question when a pointer names only part of it.

Common question typeLecture examplesDiscussionReview follow-up
Sketch a given curve5, 6, 7, 8, 9, 10, 12(i)(ii), 13(ii)1(a)–(e), 2(i)(ii), 3(i)(iii)2(i), 3(ii), 4, 5(ii)
Solve unknown constants, then sketch114, 5(i)(ii)(iv), 6(i)(iii)6(ii), 7(i)(iv), 9(ii)
Sketch involving unknowns—1(f)1
Add a second graph and count real roots12(iii)3(ii)2(ii), 4(Hence), 5(iii)
Use gradient concepts to count roots13(iii)2(iii)9(iii)
Use a discriminant to determine attainable values13(i)5(iii)3(i), 5(i), 7(iii), 9(i)
Differentiate and use a discriminant for parameter ranges—6(ii)6(i), 7(ii)

For a genuine quadratic in a real input, Δ<0\Delta<0 means no real roots (hence unattainable outputs), Δ≥0\Delta\ge0 means real roots and Δ>0\Delta>0 means two distinct real roots.

After rearranging a rational equation, check all roots against the original denominator restrictions and handle any output that makes the quadratic degenerate separately.

For derivative equations, the printed shortcuts “no stationary points: Δ<0\Delta<0” and “two turning points: Δ>0\Delta>0” require a genuine quadratic numerator, admissible roots and sign changes for turns. Cancellation and parameter boundaries can invalidate the shortcut: Review 6 retains the source’s printed k>1k>1 while explaining the literal no-stationary-point case k≥1k\ge1 and the hole at k=1k=1.

Practice questions

Practice section 1 · Questions 1–6

Find exact intercepts and verify the calculator window · Check exact intercepts of four rational curves · +4 more

Practice section 2 · Questions 7–12

Classify the two turning points exactly · Find exact input and output ranges · +4 more

Practice section 3 · Questions 13–18

Practise rational-curve features · Choose the correct branch of a modulus graph · +4 more

Practice section 4 · Questions 19–24

Use an asymptote family to find when no real roots occur · Link the discriminant interval to the missing horizontal levels · +4 more

Practice section 5 · Questions 25–30

Recast a quartic as a rational-curve intersection · Find absent output levels from a quadratic discriminant · +4 more

Practice section 6 · Questions 31–32

Use completing the square before drawing a rational graph · Recover the curve from its range and count line intersections

More practice

Before trusting a screen, predict the domain, branch behaviour and exact structural features. Explain any disagreement by checking both the rule and the window.

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