Graph transformations: shifts, reflections, absolute value

Early October 2026, Year 3 of a technical institute: after the revision of equations, inequalities, absolute value and irrational equations, in many classes the block on functions arrives (domain, composition, inverse, graphs), which comes before the parabola. The test typically falls between five and nine weeks from now, between November and December, and graph transformations are one of those questions that seem easy until you have to draw something.

The starting idea is this: if you know the graph of a function $f$, you can obtain the graph of many others without redoing the study. You only need to know what each change to the formula does to the drawing. Whoever doesn’t know redoes a table of points for every function, takes ten minutes and gets the direction of a shift wrong. Whoever knows starts from a known graph (line, parabola, $\sqrt{x}$, $1/x$), shifts, reflects, flips, and is done.

This post is precisely about drawing by hand. For the domain and symmetries of a given function there is function study, part 1; for composite and inverse functions, functions in Year 4, where transformations appear in one paragraph, whereas here we treat them in full and with more examples.

The transformations in a table

Start from $y=f(x)$. Each row says what happens to the graph:

The first four rows are the ones you always use; let’s go through them one by one.

Vertical shift: the most intuitive

$y=f(x)+b$ adds $b$ to every value of $y$, so every point $(x_0,f(x_0))$ becomes $(x_0,f(x_0)+b)$. The graph moves up or down rigidly, without changing shape. The domain doesn’t change, the range shifts by $b$. An example: from $y=x^2$ to $y=x^2-3$ the vertex goes from $(0,0)$ to $(0,-3)$.

Horizontal shift: where mistakes are most common

$y=f(x-a)$ moves the graph to the right by $a$. With $a=3$, that is $f(x-3)$, the graph goes right even though there is a minus sign; with $f(x+3)$ it goes left. At first sight it seems the opposite of what it should be, and it is mistake number one.

The way not to learn it by heart is to ask yourself: which $x$ gives me the same value of $y$ that I used to get at $x_0$? The new function $g(x)=f(x-3)$ has the same value $f(x_0)$ when the argument of $f$ is $x_0$, that is when $x-3=x_0$, so $x=x_0+3$. The value $f$ had at $x_0$, $g$ now has three units further on. The graph has slid to the right.

Example with numbers: if $f(2)=5$, then $g(5)=f(5-3)=f(2)=5$. The point $(2,5)$ has become $(5,5)$.

A consequence that gets forgotten: the domain moves too. If $f$ is defined for $x\ge 0$, $f(x-3)$ is defined for $x\ge 3$.

Reflections: $-f(x)$ and $f(-x)$

$y=-f(x)$ changes the sign of every value of $y$: the graph is reflected in the $x$-axis. Domain unchanged, zeros unchanged, points above the axis go below and vice versa.

$y=f(-x)$ changes the sign of $x$: the graph is reflected in the $y$-axis. The domain is mirrored (if $f$ is defined for $x\ge 0$, $f(-x)$ is defined for $x\le 0$), while the intercept with the $y$-axis stays put, because $-0=0$. If you apply both, $-f(-x)$, you get the reflection in the origin.

A link with what was done on function study: a function is even if $f(-x)=f(x)$, that is if its graph coincides with the mirrored one; it is odd if $f(-x)=-f(x)$, that is if reflecting in the origin doesn’t change it.

Stretches: just a mention

$kf(x)$ multiplies the heights by $k$: with $k=2$ the graph is stretched vertically (the points on the $x$-axis stay fixed). $f(kx)$ acts on $x$ and in the opposite way: with $k=2$ the graph is squeezed horizontally, because the same value of $f$ is reached with half the $x$. For example, with $f(x)=\sqrt{x}$ we have $f(4x)=\sqrt{4x}=2\sqrt{x}$: squeezing horizontally by 4 here amounts to stretching vertically by 2.

When you combine stretch and shift it is better to factorise: $f(2x-4)=f(2(x-2))$ first compresses by a factor of 2 horizontally, then moves right by 2, not by 4.

The most frequent case in tests, however, is shifts, reflections and absolute value: these are the three to be able to do with your eyes closed.

The absolute value: $|f(x)|$ versus $f(|x|)$

They are two different things and are confused constantly, so they need to be written out calmly.

$|f(x)|$: the absolute value is outside. The absolute value acts on the $y$‘s. Where $f(x)\ge 0$ the graph stays as it is; where $f(x)<0$ the value becomes $-f(x)$, so that part is flipped above the $x$-axis. The result never has points below the $x$-axis, and at the points where the graph crosses the axis corners generally appear.

$f(|x|)$: the absolute value is inside. The absolute value acts on the $x$‘s. For $x\ge 0$ we have $|x|=x$ and the graph is that of $f$. For $x<0$ we have $|x|=-x$, so $f(|x|)=f(-x)$: the graph on the left is the mirror image in the $y$-axis of the one on the right. The part of $f$ that was on the left (for $x<0$) disappears: it is replaced by the mirror image of the right-hand part. The result is always an even function.

Practical rule so as not to swap them: look at where the absolute value is written. Outside, on the $y$: flip what is below. Inside, on the $x$: copy the right-hand side to the left.

A halfway check

At this point try drawing by hand, without looking at the examples below, $y=\sqrt{x-1}+3$ and $y=-(x+2{)}^2+4$. To train against the clock, the practice tests (15 questions in 50 minutes, one per topic) are at /simulazioni/verifiche-liceo/.

How to draw by hand

1. Recognise the base function and draw it, marking 3-4 key points: vertex, intercepts with the axes, any asymptotes.
2. Apply the transformations one at a time, from the inside out (first what acts on $x$, then what acts on $y$).
3. Transform the key points, not just the shape: write down where the vertex, the zeros and the intercept end up.
4. Read off the domain and range from the drawing and check them on a point.

If the base function is a parabola, the same steps become those of the chapter on the parabola: vertex, zeros, axis. It is worth arriving there with the transformations already in hand.

Three worked exercises

1. $y=x^2-4x+3$ and its absolute value

You are asked for: vertex, intercepts, graph of $y=|x^2-4x+3|$.

Completing the square:

so the parabola is $y=x^2$ shifted right by 2 and down by 1. Vertex $V(2,-1)$. Zeros: $x^2-4x+3=0$ gives $(x-1)(x-3)=0$, that is $x=1$ and $x=3$. Intercept with the $y$-axis: $(0,3)$. A fifth point symmetric about the axis $x=2$: $(4,3)$.

The function is negative only between the zeros, for $1<x<3$. For $|f(x)|$ exactly that piece is flipped: the vertex becomes $(2,1)$, the zeros stay at $(1,0)$ and $(3,0)$, the rest doesn’t change. In formulas:

Check on the vertex: for $x=2$ the second branch gives $-4+8-3=1$, consistent with $(2,1)$. The graph has two corners, at $(1,0)$ and $(3,0)$, and between the two the flipped parabola is concave down.

To see what would change with the absolute value inside, the graph of $f(|x|)=x^2-4|x|+3$ is built like this: to the right of the $y$-axis the original parabola stays, with $(0,3)$, $(1,0)$, $(2,-1)$, $(3,0)$; to the left these points are mirrored: $(-1,0)$, $(-2,-1)$, $(-3,0)$. It is a symmetric “W”, with two minima at $(\pm 2,-1)$. The original piece to the left of the $y$-axis (the branch that came down from $(0,3)$ towards the vertex) no longer appears: it has been erased and replaced by the mirror image of the right-hand one.

A question that links to equations: how many solutions does $|x^2-4x+3|=1$ have? It splits into $x^2-4x+3=1$, that is $x^2-4x+2=0$, with $x=2\pm \sqrt{2}$, and into $x^2-4x+3=-1$, that is $x^2-4x+4=0$, with $x=2$ a double root. The solutions are $2-\sqrt{2}$, $2$, $2+\sqrt{2}$: three intersections with the line $y=1$, and they can be seen straight away in the drawing, because the line touches the flipped vertex and cuts the two outer branches.

2. From $y=\sqrt{x}$ to $y=\sqrt{x+2}-1$

The base graph starts at $(0,0)$ and rises: domain $x\ge 0$, range $y\ge 0$, points $(1,1)$, $(4,2)$.

$\sqrt{x+2}$ is a horizontal shift to the left by 2; the final $-1$ moves down by 1. The starting point $(0,0)$ goes to $(-2,-1)$. Domain: $x+2\ge 0$, that is $x\ge -2$. Range: $y\ge -1$.

Points to plot, obtained by calculating directly and checking:

• $x=-2$: $\sqrt{0}-1=-1$, point $(-2,-1)$;
• $x=-1$: $\sqrt{1}-1=0$, point $(-1,0)$, which is the intercept with the $x$-axis;
• $x=0$: $\sqrt{2}-1\approx 0.41$, intercept with the $y$-axis;
• $x=2$: $\sqrt{4}-1=1$, point $(2,1)$;
• $x=7$: $\sqrt{9}-1=2$, point $(7,2)$.

Consistency check: the base points $(1,1)$ and $(4,2)$ shifted by $(-2,-1)$ give $(-1,0)$ and $(2,1)$, which are exactly the ones found by calculating.

3. From $y=1/x$ to $y=\frac{1}{x-1}+2$

The hyperbola $y=1/x$ has asymptotes $x=0$ and $y=0$. The shift of $+1$ in $x$ and of $+2$ in $y$ moves everything, asymptotes included:

Domain $x\ne 1$, range $y\ne 2$. You don’t need a table of points to know where the branches are: for $x>1$ the term $\frac{1}{x-1}$ is positive, so the branch lies above $y=2$; for $x<1$ it is negative, so the branch lies below.

Intercepts: with the $y$-axis, $x=0$ gives $-1+2=1$, point $(0,1)$. With the $x$-axis: $\frac{1}{x-1}=-2$, so $x-1=-\frac{1}{2}$, $x=\frac{1}{2}$, point $\left(\frac{1}{2},0\right)$. Other points: $x=2$ gives $3$, $x=3$ gives $\frac{5}{2}$, $x=-1$ gives $\frac{3}{2}$.

If the same function is given to you as $y=\frac{2x-1}{x-1}$ (a homographic function, i.e. a ratio of two linear expressions), you recognise it by rewriting it: $\frac{2x-1}{x-1}=\frac{1}{x-1}+2$, and so it is the same hyperbola shifted. The vertical asymptote is where the denominator vanishes, the horizontal one is the ratio of the coefficients of $x$.

Typical mistakes

• The sign in the horizontal shift. $f(x-3)$ goes right. Check with a point: if $f(2)=5$, the new graph passes through $(5,5)$.
• The absolute value on $x$ instead of on $y$, or vice versa: look at where it sits and apply the inside/outside rule.
• Confusing $|f(x)|$ and $f(|x|)$. The first flips what is below, the second replaces the left-hand part with the mirror image of the right. In example 1 the two functions have very different drawings.
• Forgetting that the domain moves together with the graph: in $\sqrt{x+2}-1$ the domain is $x\ge -2$, not $x\ge 0$.
• Applying the transformations in the wrong order, in particular with stretch and shift together: factorise and read from the inside out.

Link with equations and inequalities with absolute values

Drawing is the quickest way to check an equation or inequality with an absolute value. When you solve $|x^2-4x+3|=1$ algebraically, you split it into two cases; the graph tells you how many results to expect and where they lie. The inequality $|x^2-4x+3|<1$ reads as “the part of the graph of $|f|$ below the line $y=1$”. The algebraic rules, with the conditions to be written first, are in absolute value and irrational equations in Year 3; here the drawing is enough to see whether the result is plausible.

To practise

The practice tests in the catalogue are 15 questions in 50 minutes, one per topic, and are a good way of seeing whether the graph part holds up against the clock: you will find them at /simulazioni/verifiche-liceo/. There is no dedicated test on graph transformations alone: in the index you will find the topics that already exist.

How we work in lessons

I always start from the drawing: whoever arrives at this chapter in difficulty usually doesn’t have a problem with formulas, but doesn’t yet have the reflex of transforming the key points. We work on one base function at a time, with one transformation at a time, and only afterwards combine them. I have you draw two or three graphs by hand on paper in the lesson and then ask you to explain to me out loud why the direction is what it is. If the test is still far away we build the method; if it is close we work directly on typical papers. The subject is maths; for details on the format of lessons there is the pricing page.

If you want to understand where to restart, the free 30-minute introductory call is a short conversation in which we look together at a piece of homework or a notebook and decide whether you really need a path or a couple of tips are enough.