How to Improve Additional Mathematics | Function Graph Transformations and Point Mappings

Quick Read: Transform Points Before Sketching the Curve

Improving function graph transformations in Additional Mathematics begins with understanding where familiar points move when a function rule changes. If (a,b) is on y=f(x), each transformation has a definite effect on the coordinates. A graph shifted three units right sends that point to (a+3,b); one shifted two units up sends it to (a,b+2). Using point mappings makes it possible to verify a sketch mathematically rather than relying entirely on remembered directions or visual guesses.

The greatest source of confusion is the difference between changes outside f and changes inside its input. In y=f(x)+3, adding three happens to the output, so the graph moves up. In y=f(x+3), adding three happens before the function receives its input, and the graph moves left. The inside transformation appears to work in the opposite direction because the new input must compensate for the added number to recreate the original output.

Students should begin with a known point, transform it and substitute its new coordinates into the proposed equation. They should then check how the vertex, endpoints, asymptotes, domain and range move. These habits make graphing more accurate and connect to function composition, algebraic inequalities, coordinate geometry and calculus. Not every extension transformation discussed here is necessarily a separately assessed item in the G3 K341 curriculum; check the published syllabus for the student’s own examination year.

What Inside and Outside Operations Mean

Let f(2)=5. Under y=f(x)+4, the input remains 2 and the output becomes 9, giving (2,9). Under y=f(x−4), to reproduce the old input 2, set x−4=2 and obtain x=6. Thus the corresponding point is (6,5). This two-line argument is more reliable than trying to remember an arbitrary slogan about plus and minus signs.

For a vertical stretch y=3f(x), the old point (2,5) moves to (2,15). For a horizontal compression y=f(3x), it moves to (2/3,5). In the first expression the output is multiplied by three. In the second, the input supplied to f is three times the displayed x, so the displayed x must be one-third of the old input. The algebra explains why the transformations differ.

If the child cannot explain which quantity is being changed, a large worksheet is unlikely to resolve the uncertainty. Instead, practise a few numerical input-output pairs and require brackets around substituted expressions. The basic skill is recognising that f is a rule operating on its entire input. Once that meaning is secure, the more abstract idea of transforming a whole curve becomes easier.

Horizontal Translation Is an Input Equation

Start with y=x², whose vertex is (0,0). For y=(x−3)², the squared expression becomes zero when x=3, so the vertex is (3,0). For y=(x+3)², it becomes zero when x=−3, giving vertex (−3,0). The new equations provide immediate checks for the direction of the shift. A minus inside the input produces a rightward shift in this example, while a plus produces a leftward shift.

In general, for y=f(x−h), an old point (a,b) appears where x−h=a, so the new point is (a+h,b). For f(x+h), solve x+h=a and obtain (a−h,b). This works for any function on the appropriate domain, not just quadratics. If an old point is excluded from the domain, do not pretend it lies on the graph simply to apply the mapping.

A student can practise with a reciprocal graph f(x)=1/x. Its vertical asymptote is x=0. For 1/(x−3) it becomes x=3 because that is the forbidden input. The branches move with the shift. Asymptotes are part of the graph’s mathematical structure, not decorations added at the end.

Vertical Translation Changes the Output Directly

For y=f(x)+k, an old point (a,b) becomes (a,b+k). The x-coordinate does not change, because the function still receives the same input. A graph with a maximum y-value of 5 has maximum y-value 8 after an upward translation by three, provided the same original domain applies. Likewise, a horizontal asymptote at y=0 moves to y=3 if three is added outside the function.

The input domain is unchanged by a simple vertical shift, but the range shifts. The parent square-root function y=√x has range y≥0. Its translation √x+4 has range y≥4 while retaining domain x≥0. An outside negative shift would lower every output by the stated amount. Students who change the domain when only an outside constant is added have confused input and output roles.

The First Point-Mapping Reference

Transformation	Mapped point (a,b)	Graph effect
`f(x−3)`	(a+3,b)	Right 3
`f(x+3)`	(a−3,b)	Left 3
`f(x)+2`	(a,b+2)	Up 2
`2f(x)`	(a,2b)	Vertical stretch 2
`f(2x)`	(a/2,b)	Horizontal compression 2
`−f(x)`	(a,−b)	Reflect in x-axis
`f(−x)`	(−a,b)	Reflect in y-axis

The table is a guide to building understanding, not a substitute for it. Students should be able to prove each coordinate change by substituting into the transformed rule. When the parent graph is symmetric, a reflection may appear to leave its shape unchanged. Use a nonsymmetric parent function such as f(x)=x+1 or f(x)=√x to make the distinction between the axes of reflection visible.

Vertical Scaling and Reflection in the x-Axis

For a vertical scale y=Af(x), an original point (a,b) maps to (a,Ab). If A=3, each y-coordinate triples and the graph stretches away from the x-axis; if A=1/2, the vertical distances halve. If A is negative, the outputs change sign and the graph reflects in the x-axis as well as being scaled by |A|. The input x remains unchanged throughout. This differs from changing the scale inside the function’s argument.

Take the parent function f(x)=x². The point (2,4) becomes (2,12) on y=3x², and (2,−8) on y=−2x². The vertex (0,0) stays fixed because zero multiplied by a constant remains zero. The negative version opens downwards, which is visible from its transformed points. Students should not confuse this with reflection in the y-axis; the parabola is symmetric and would appear unchanged under that different reflection.

The order of outside operations matters. The graph y=−f(x)+2 reflects the old outputs, then adds two. The graph y=−[f(x)+2] negates both the old output and the added two. They are not equivalent. At a point where f(a)=5, the first gives −3 and the second gives −7. This numerical comparison is a practical way to identify misplaced brackets.

Horizontal Scaling Changes Old x-Coordinates

For y=f(Bx) with B positive, an old point (a,b) is reached when Bx=a. The transformed point is therefore (a/B,b). A factor B=2 inside compresses the graph horizontally by two, while B=1/2 stretches it horizontally by two. These statements describe where the original features occur on the new horizontal axis, not how fast an individual point physically moves.

Suppose the parent graph contains (3,9). On y=f(2x) the corresponding point is (1.5,9) because the inner input at x=1.5 is 3. On y=f(x/3) it is (9,9) because the new input must be divided by three to equal 3. A learner can derive both positions without sketching a curve first. This is the most reliable way to avoid reversing a compression and a stretch.

This same idea explains periods in trigonometry. The parent sine wave repeats after 360°; replacing x with 2x makes the inner angle complete a revolution over just 180° of the displayed x-axis. Replacing x with x/2 makes the cycle take 720°. Connecting point mappings with periods gives students one underlying principle rather than two unrelated rules to remember.

A negative inside factor also reflects horizontally. For y=f(−x), the old point (a,b) maps to (−a,b). On an unsymmetric function such as f(x)=x+1, the original point (2,3) becomes (−2,3). By contrast, y=−f(x) maps it to (2,−3). Always identify which coordinate the negative sign affects.

Combined Transformations: One General Point Rule

Consider y=A f(B(x−h))+k where A and B are nonzero. To obtain the old function output b=f(a), we require B(x−h)=a. Solving gives x=h+a/B. The new output is Ab+k. Therefore an old point (a,b) maps to (h+a/B,Ab+k). This formula combines horizontal and vertical scales, reflections and translations without relying on a fragile verbal order of operations.

For example, suppose f(6)=5 and the new function is y=−2f(3(x−1))+4. Solve the input equation 3(x−1)=6 to get x=3. Apply the output rule: −2(5)+4=−6. Thus (6,5) maps to (3,−6). Substitute x=3 back into the transformed function to confirm that its inside argument is 6 and its final output is −6.

The same method can be applied to every point of a parent graph. A student does not have to memorise whether “stretch then translate” differs from “translate then stretch” in the teacher’s chosen description. The algebra of the complete function determines the mapping. However, the result still has to respect the original domain: a point that was never in the parent function cannot be introduced as if it were an ordinary mapped point.

Worked Example 1: A Shifted and Reflected Quadratic

Starting from f(x)=x², consider y=−2(x−3)²+4. The old point (a,a²) maps to (a+3,4−2a²). The vertex (0,0) becomes (3,4). Since the leading multiplier is negative, the new parabola opens downwards and has a maximum y-value of 4. The domain remains all real numbers, while its range becomes y≤4.

The original symmetric points (−1,1) and (1,1) map to (2,2) and (4,2). Likewise (−2,4) and (2,4) map to (1,−4) and (5,−4). Plotting these points around the new vertical axis of symmetry x=3 produces a reliable sketch. If the student’s points are not symmetric around x=3, an x-coordinate mapping or arithmetic error has occurred.

Find the x-intercepts by setting the transformed y-value to zero: −2(x−3)²+4=0, so (x−3)²=2 and x=3±√2. The y-intercept occurs at x=0 and is −2(−3)²+4=−14. A graph transformation gives the vertex immediately, but intercept questions can still require exact algebra. A student should select the correct tool for the requested result.

There is a useful self-check: the two x-intercepts are equally spaced around x=3, matching the graph’s symmetry. Their average is 3. If the calculated roots do not share that midpoint, inspect the expansion or square-root step. The sketch and the algebra should support each other, not operate as competing ways to answer the problem.

Worked Example 2: Reciprocal Curves and Asymptotes

Take f(x)=1/x, defined for x≠0, and consider g(x)=2/(x−3)+1. The graph shifts right three, scales vertically by two and moves up one. Its vertical asymptote is x=3 because the denominator becomes zero there. Its horizontal asymptote is y=1 because the reciprocal term approaches zero for large positive or negative x. These lines move with the graph’s structure.

The parent points (1,1), (2,1/2), (−1,−1) and (−2,−1/2) map to (4,3), (5,2), (2,−1) and (1,0). Check one directly: at x=1, g(1)=2/(−2)+1=0. This also identifies the x-intercept. At x=4, g(4)=3. These values help place both reciprocal branches without guesswork.

The transformed domain is all real x except 3, and its range is all real y except 1. It would be mathematically incorrect to draw the curve crossing either asymptote or to label the domain as x≠0 copied from the parent function. The input and output restrictions are part of the transformation. A clear graph displays them explicitly.

Worked Example 3: Square Roots and the Starting Point

Let f(x)=√x, defined for x≥0, and transform it into g(x)=−3√(x−4)+2. The inside shift moves the parent starting point (0,0) to x=4. The negative outside factor reflects the outputs and stretches their magnitudes by three; the final +2 moves them upwards. The new graph begins at (4,2) and decreases as x increases.

The domain requires x−4≥0, so x≥4. Since the square-root value is nonnegative, multiplying by −3 produces a nonpositive quantity and adding 2 gives range y≤2. The original points (0,0), (1,1), (4,2) and (9,3) map to (4,2), (5,−1), (8,−4) and (13,−7). These give a consistent sequence of coordinates for the curve.

If a child keeps the original domain x≥0 after shifting the radical, the error is conceptual rather than artistic. A transformed formula must be evaluated at its own allowed inputs. Writing the inequality for the radicand before sketching is an efficient repair and connects this topic with the wider algebraic practice of preserving the domain of a function.

Domains, Ranges and Removed Points

Under an ordinary vertical shift or nonzero vertical scale, the input domain of a function stays the same. A horizontal translation or scaling changes the input coordinates and therefore changes its domain. The range changes under output operations. A student should not merely copy old restrictions into the new answer. Instead, transform the original domain and range or derive the restrictions directly from the formula.

For example, √(x−3) has real domain x≥3; √x−3 has domain x≥0 but range y≥−3. The outside and inside positions of the same number produce different restrictions. Likewise, the reciprocal functions 1/(x−3) and 1/x−3 have different asymptotes and domains. A pair of numerical test inputs can make the difference immediately visible.

A rational expression may simplify to a formula that appears defined at an originally forbidden input. That does not automatically repair the original function’s domain. For example, (x²−1)/(x−1) simplifies algebraically to x+1 for x≠1. The original has a hole at x=1, even though x+1 is defined there. Students should preserve the original exclusion in the graph and in any domain statement. This principle carries over to composite functions and logarithms.

Inverse Functions Are Reflected in a Different Line

An inverse function should not be confused with the reflection transformations −f(x) and f(−x). When a one-to-one function has an inverse, its graph reflects in the diagonal line y=x, swapping each original point (a,b) to (b,a). That is fundamentally different from reflecting in the horizontal x-axis or vertical y-axis. If f(x)=2x+3, the original point (1,5) corresponds to inverse point (5,1).

The inverse of f(x)=2x+3 is f⁻¹(x)=(x−3)/2. Substitution verifies that f(1)=5 and f⁻¹(5)=1. The input and output roles exchange, so the inverse’s domain equals the original function’s range and vice versa. A student who simply writes −f(x) when asked for an inverse has confused a reflection of outputs with a reversal of the mapping.

The one-to-one condition also matters. For f(x)=x² over all real numbers, both x=2 and x=−2 produce y=4. Reflecting the entire parabola across y=x gives a sideways relation with two outputs for many inputs. Restricting the original domain to x≥0 selects an inverse f⁻¹(x)=√x. The graph and domain conditions together explain why a single inverse branch is necessary.

Absolute-Value Graphs as a Useful Extension

A student who has mastered translations and reflections can investigate y=|f(x)| and y=f(|x|) as extension examples. In the first, negative outputs are reflected upwards while nonnegative outputs stay unchanged. In the second, the input is replaced by its absolute value, so the function’s right-hand behaviour is mirrored into the left half-plane. These transformations illustrate input–output reasoning even where a particular syllabus does not assess them separately.

Take f(x)=x−2. The outside absolute value gives |f(x)|=|x−2|, whose V-shaped graph has vertex (2,0). The inside absolute value gives f(|x|)=|x|−2, which has vertex (0,−2). Both expressions contain the same symbols but yield different graphs. A student who can explain their difference has understood the location of an operation within a function.

These cases also teach piecewise reasoning. To draw |f(x)|, keep the parts of f that are already above the x-axis and reflect any portions below it. To draw f(|x|), use the original graph for nonnegative inputs and reflect that right-hand portion across the y-axis. The technique should be justified through the definition of absolute value, not memorised as an unexplained sketching trick.

How Transformations Affect a Derivative

Graph transformations also connect to calculus through the chain rule. For a differentiable function f, if g(x)=A f(B(x−h))+k with A and B constant, then g′(x)=AB f′(B(x−h)). The vertical shift k differentiates to zero, while both the outside vertical scale and the inside horizontal factor influence the new gradient. This connection provides a mathematical check on the apparent steepness of the transformed graph.

For example, set f(x)=x² and g(x)=3f(2x). The expanded expression is g(x)=12x², so g′(x)=24x. The chain-rule route gives 3×2×f′(2x)=6×(4x)=24x, in agreement. If a learner differentiates the outside 3 correctly but forgets the factor 2 from the inside input, the derivative will be wrong even when the transformed graph has been sketched correctly.

For g(x)=−2(x−3)²+4, the derivative is −4(x−3), which vanishes at x=3, the transformed vertex. Its sign changes from positive to negative there, matching the downward-opening parabola. The graph makes the stationary-point answer plausible while differentiation supplies a formal gradient calculation. Connecting the two helps students check work across chapters.

A Method-Choice Checklist Before Sketching

When a new function appears, first identify the parent function and its original domain. Next circle operations outside f, which change outputs, and underline operations inside its argument, which change inputs. Then choose at least three anchor points and map them using the appropriate coordinate rule. Finally determine the transformed domain, range, turning point or asymptotes, and substitute one mapped point into the new equation. This sequence replaces guesswork with a reproducible method.

The number of points needed depends on the curve. A quadratic is strongly constrained by its vertex and symmetry, while a reciprocal requires attention to branches and asymptotes. A square-root graph has an endpoint and a one-sided domain. A trigonometric graph requires enough landmarks to show the period. The student should choose points that reveal structure rather than simply plot many arbitrary decimals.

If a question asks for an exact intercept, mapping old points may not be enough. Solve the new equation with y=0 or x=0 as appropriate. If it asks for a graph inequality, identify where the curve lies above or below an axis. If it asks for an inverse, investigate one-to-one behaviour rather than merely shifting a curve. A reliable learner uses transformations as one tool within the larger function-solving system.

A Summary of Transformations and Checks

Expression	Main effect	Best checking habit
f(x−h)	Right shift by h	Solve x−h=a
f(x)+k	Vertical shift by k	Check new y=b+k
Af(x)	Vertical scale; x-axis reflection if A<0	Multiply a known output
f(Bx)	Horizontal scale by 1/\|B\|; y-axis reflection if B<0	Solve Bx=a
−f(x)	Reflect in x-axis	Map (a,b) to (a,−b)
f(−x)	Reflect in y-axis	Map (a,b) to (−a,b)
f⁻¹(x)	Reflect in y=x, when inverse exists	Swap coordinates and check f(f⁻¹(x))

The point-checking habit is the same across the table: start with an input and output you know are valid. Transform them explicitly, then test the result in the new equation. Students need not redraw a full graph for every small check, but they should be able to justify their proposed shift or reflection without relying on an unexamined phrase such as “inside signs are backwards”.

Why Some Graph Mistakes Are Actually Algebra Mistakes

If a student knows a parabola should move right by three yet writes (x−3)²=x²−9, the transformation concept may be correct but binomial expansion is weak. The valid expansion is x²−6x+9. A tutoring plan should repair the distributive law and squared-binomial identity, not reteach the horizontal shift. This distinction can prevent wasted repetition and frustration.

Other students factorise and substitute accurately yet draw the wrong graph because they treat f(2x) like 2f(x). Their issue is interpreting function notation rather than ordinary algebra. One student’s correction might be three short input equations; another’s might be five examples expanding brackets. A common low test score can conceal very different learning needs.

A useful written correction record contains the original instruction, the earliest incorrect line, the reason it fails and a replacement question using different values. On the following day, the student should attempt the replacement without seeing the corrected solution. Success after a short delay is stronger evidence of understanding than repeating the identical graph while the teacher’s method remains in view.

A Two-Minute Diagnostic for Parents and Tutors

Ask the learner to start from a point (2,5). Where does it move under f(x−3)? Under f(x)+3? Under −f(x)? Under f(2x)? A student who answers (5,5), (2,8), (2,−5) and (1,5) and can explain each answer has most of the central input–output distinction. A student who gives the same movement for the first two transformations needs a simpler explanation before moving to complete function sketches.

Then offer a parent graph with a visible restriction, such as f(x)=√x, and ask about the domain of √(x−4). If the learner says x≥4 and can explain that x−4 must be nonnegative, the domain concept is emerging. If they simply copy x≥0, the restrictions have not transferred with the graph. These two short exercises reveal important information without requiring an entire mock examination.

A Fourteen-Day Plan for Better Graph Transformations

Days 1–2: Make Function Notation Clear

Begin with a simple rule such as f(x)=2x+1 and practise finding f(3), f(a+1), f(2x) and 2f(x). Keep brackets visible when substituting expressions. Ask the student to explain why changing an input differs from multiplying an output. The aim is to establish that f is a rule acting on the entire input, rather than a symbol being multiplied by x.

Days 3–4: Work with Two Kinds of Shift

Use y=x² and y=√x as parent curves. Shift each by three units horizontally and then by three vertically, using two separate questions. Map at least three known points each time and explain the different changes. Ask the student to compare f(x+3) with f(x)+3 and confirm by direct substitution. This contrast is the most important early repair for graph-translation confusion.

Days 5–6: Add Scaling and Reflection

Practise 2f(x), f(2x), −f(x) and f(−x) on a nonsymmetric parent function. Use a line or square-root curve, not only an even quadratic, so that reflections visibly differ. Record the mapped points in a compact table. End each session by explaining why an inside factor divides old x-coordinates while an outside factor multiplies old y-coordinates.

Days 7–8: Combine Two Transformations

Introduce a quadratic containing one horizontal shift and one vertical scale, then add a vertical shift. Map the vertex and symmetric points, state the domain and range, and use substitution to check a coordinate. Only after those stages are stable should the student attempt the general form Af(B(x−h))+k. Avoid rushing into four operations while basic sign rules remain uncertain.

Days 9–10: Rational and Square-Root Domains

Use reciprocal functions with moving vertical and horizontal asymptotes and square-root functions with moving endpoints. Ask for the forbidden input of a denominator or the nonnegative condition on a radicand before the sketch is drawn. A correct graph should respect these mathematical restrictions. If a student copies the parent domain unchanged, practise deriving the condition from the transformed formula.

Days 11–12: Link Transformations to Other Topics

Connect quadratic transformations with completing the square, trigonometric scaling with period, and inverse functions with reflection in y=x. Give one simple derivative of a transformed function and confirm it using the chain rule. These connections make the technique a useful mathematical language instead of an isolated graphical chapter. Keep the exercises appropriate to the student’s actual syllabus and readiness.

Days 13–14: Unseen Mixed Questions

Remove method labels and present several transformed functions with varied numbers and signs. The student should identify the parent, map anchor points, state domain and range where relevant, and check at least one coordinate. Record the first invalid step and retest that error category after a delay. A fresh correct graph with an explanation is more meaningful than a neat tracing of a worked example.

Independent Self-Test with Answer Checks

1. A parent graph contains (2,5). Under f(x−3), it moves to (5,5).

2. Under f(x)+3, the same point moves to (2,8).

3. Under −f(x), it moves to (2,−5).

4. Under f(2x), it moves to (1,5).

5. Under f(x/2), it moves to (4,5).

6. Under f(−x), it moves to (−2,5).

7. The vertex of y=(x−4)²−3 is (4,−3).

8. The domain of y=√(x−3)+2 is x≥3.

9. The range of y=−2√(x−3)+2 is y≤2.

10. The vertical asymptote of y=2/(x−5)+1 is x=5.

11. The horizontal asymptote of the same function is y=1.

12. Under y=−2f(3(x−1))+4, the old point (6,5) moves to (3,−6).

13. For a genuine inverse, the point (a,b) on f becomes (b,a) on f⁻¹.

14. Explain why f(x−3) and f(x)+3 have different domains when f is the square-root function.

For each item, ask for one sentence explaining the coordinate or restriction. If a student gets the right answer but cannot explain the inside scaling, the skill may still be fragile. If the student can correctly map points in an unseen combined transformation, identify exceptional inputs and describe the curve, the method is becoming independently transferable.

How Parents Can Interpret a Graphing Error

Parents do not need to reproduce a complete graph to ask useful diagnostic questions. Start with “Which point did you know on the old curve?” and “Where must it go in the new equation?” The child should be able to show one substitution that confirms the answer. If they cannot, the difficulty may be function notation rather than artistic sketching. Focus the next practice on the earliest confusion that appears.

A child who understands horizontal and vertical shifts but repeatedly expands (x−3)² incorrectly needs algebra practice. A child who expands accurately yet places the vertex at x=−3 needs point-mapping practice. Another who gets the vertex right but includes an undefined reciprocal input needs domain practice. These are different problems despite their appearance in the same graph question. Targeted teaching is more useful than repeatedly assigning the entire topic.

Parents can support regular short attempts, clear corrections and delayed retesting. A correct solution copied immediately after a tutor’s example is encouraging, but it is not the same as independent understanding. The stronger indicator is being able to explain a changed function and produce a new graph several days later without prompts.

How Small-Group Tutorials Can Repair the Right Skill

In eduKateSG’s three-student tutorial format, a tutor can watch the first coordinate mapping before the full graph appears. One learner may confuse inside and outside operations, another may omit a domain restriction and a third may be ready to combine transformations with differentiation. They do not need identical worksheets simply because the class is studying functions. Close inspection supports a more precise next teaching step.

The sequence is diagnose, repair prerequisites, practise accurately, connect with other chapters, test on unfamiliar questions, correct and retest. The objective is for students to understand why a point moves and what remains mathematically valid, not to memorise an impressive number of finished sketches. Independent method choice and justified restrictions matter as much as the appearance of the final curve.

Frequently Asked Questions

Why does f(x−3) move right, not left?

To reproduce the original value f(a), the new displayed input x must satisfy x−3=a. Hence x=a+3. Every old point moves three units to the right. Solving this input equation is a reliable explanation that works with shifted square roots, quadratics, reciprocal functions and other valid parent functions.

What is the difference between 2f(x) and f(2x)?

The first multiplies each output by two and creates a vertical stretch. The second multiplies the input sent into f by two, so old x-coordinates are halved and the graph is horizontally compressed. Use a known point (a,b) to verify that they produce (a,2b) and (a/2,b) respectively.

Do transformations change domains and ranges?

Horizontal transformations can alter which inputs are allowed; vertical transformations can change the output range. Square-root and reciprocal restrictions move with the formula, while a simple outside vertical translation leaves the input domain unchanged. Always check denominators, radicands and any original exclusions after transforming the equation.

Is finding an inverse just another x-axis reflection?

No. An inverse function reflects the graph across y=x and swaps input and output coordinates. Reflection in the x-axis negates outputs. A function must also be one-to-one on the chosen domain to have an inverse function. These operations serve different mathematical purposes.

How can my child check a transformed graph without software?

Choose two or three parent points and map them using the transformation rule. Substitute the new coordinates into the new equation and check special features such as a vertex, starting point or asymptote. For an intercept or domain question, use the original algebraic condition as an independent check. A verified rough sketch is more useful than a detailed but unjustified one.