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A JEE question asks how many real solutions |x² − 4x + 3| = k has. Solving algebraically takes several cases. Sketching the graph takes a minute, if you can transform graphs quickly and correctly. This post gives you the rules, a safe order for applying them, worked examples and the mistakes that cost marks.
Why do graph transformations matter for JEE?
The JEE (Main) 2026 syllabus asks for "graphs of simple functions" in Unit 7, on limits, continuity and differentiability, and graphs are the fastest route to answers about the number of solutions, the range, and where a function is not differentiable. The NCERT Class 11 chapter on relations and functions gives the basic graphs you build from, including the modulus, signum and greatest integer functions. Check the syllabus for your exam year.
What are the basic transformations?
Start from y = f(x):
y = f(x) + b: shift up by b (down if b is negative).
y = f(x − a): shift right by a (left if a is negative).
y = a f(x): stretch vertically by a factor a.
y = f(ax): squeeze horizontally by a factor a, for a > 1.
y = −f(x): reflect in the x-axis.
y = f(−x): reflect in the y-axis.
y = |f(x)|: keep the part above the x-axis and reflect the part below it upwards.
y = f(|x|): keep the part for x ≥ 0, delete the left part, and mirror the right part into the left.
Why does the horizontal shift feel backwards?
Almost everyone expects y = (x − 3)² to sit three units to the left of y = x². It sits three units to the right. Zazkis, Liljedahl and Gadowsky (2003) asked secondary students and teachers to explain this exact example and found the horizontal shift was, at least at first, counterintuitive to most of them. Their explanations relied on spotting patterns, on locating where the zero of the function went, or on calculating values point by point.
That gives you a reliable check. For y = (x − 3)², the input must be 3 to give the output that x² gives at 0. So the vertex moves from x = 0 to x = 3. Test one point whenever you are unsure.
Worked examples
Worked example 1. Sketch y = −2(x + 1)² + 3.
Start with y = x².
Replace x by x + 1: shift left by 1, vertex at (−1, 0).
Multiply by 2: narrower parabola.
Multiply by −1: opens downwards.
Add 3: vertex at (−1, 3). Check: at x = 0, y = −2 + 3 = 1, and the graph should pass through (0, 1).
Worked example 2. How many real solutions does |x² − 4x + 3| = k have?
Complete the square: x² − 4x + 3 = (x − 2)² − 1. Vertex (2, −1), roots x = 1 and x = 3.
Apply the modulus: the dip between x = 1 and x = 3 flips upwards, making a hump with its top at (2, 1).
Draw horizontal lines y = k and count crossings. For k < 0 there are none. For k = 0 there are two (x = 1 and x = 3). For 0 < k < 1 there are four. For k = 1 there are three, because the line touches the top of the hump at x = 2. For k > 1 there are two.
Check k = 1 by algebra: (x − 2)² − 1 = ±1 gives (x − 2)² = 2 (two roots) or (x − 2)² = 0 (one root), three in total.
Worked example 3. Solve ||x| − 2| = 1 using the graph.
y = |x| is a V with its corner at the origin.
y = |x| − 2 moves it down by 2, crossing the axis at x = ±2.
Applying the outer modulus flips the part between −2 and 2 upwards, giving a W shape with y = 2 at x = 0.
The line y = 1 cuts the W four times. By algebra, |x| − 2 = ±1 gives |x| = 3 or |x| = 1, so x = ±3 or ±1.
Every corner in these graphs is a point where the function is not differentiable. Our post on differentiability for JEE shows how to check corners like these.
In what order should you apply transformations?
Inside changes (to x) act horizontally; outside changes (to f) act vertically. A safe order is: inside shifts and stretches first, then vertical stretches and reflections, then vertical shifts, and modulus last unless the modulus is inside.
The trap is a combined inside change such as y = f(2x − 6). Write it as f(2(x − 3)): squeeze by 2, then shift right by 3, not 6. Check with f(x) = x²: y = (2x − 6)² = 4(x − 3)², vertex at x = 3.
Common mistakes and how to avoid them
Shifting left for f(x − a). Test one point.
Shifting by 6 instead of 3 in f(2x − 6). Factor the coefficient out first.
Confusing |f(x)| with f(|x|). The first folds the bottom up; the second copies the right half to the left.
Forgetting the touching case when counting solutions, like k = 1 in worked example 2.
Trusting a graphing app's picture without understanding it. Our post on using graphing tools to learn functions shows how to use them as a check, not a crutch.
A practice routine
Days 1 to 2: Ten one-step transformations of x², |x|, √x, 1/x and sin x. Sketch, then test one point.
Days 3 to 4: Ten two- and three-step transformations, including f(ax + b) forms.
Days 5 to 6: Ten "number of solutions" questions using modulus graphs, with an algebra check for the touching cases.
Day 7: A timed mixed set of eight.
If a basic graph such as the greatest integer function is still fuzzy, the Fermi Concept Library covers every concept in mathematics with its own lesson, so you can revise the starting graph before transforming it. For functions whose domain causes trouble, see our guide to domain and range.
Start now: sketch y = |x² − 4x + 3| on paper and mark the four crossing points for k = 1/2.
