A calculus problem almost never hands you a clean formula to work with in isolation. More often it hands you a graph, or a shifted and stretched version of something you already know. It expects you to read that fluently. The fluency is worth refreshing now, before it's the bottleneck in a problem about something new.
Reading a graph fluently means your eye pulls out facts a formula would take real algebra to find. Where does the curve cross zero? Is it climbing or falling right here? Where's it headed way off to either side? Picture the function's graph as a photograph you can drag around and resize. Sliding the whole photo left, right, up, or down is a shift. Stretching it taller or squishing it flatter is a vertical scale. One move is counterintuitive: anything that happens to x itself before the function sees it, like f(x-3). You aren't moving the photo there. You're relabeling the ruler underneath it. Relabeling "3" to mean what "0" used to mean pushes everything the other direction.
Reading a graph means extracting facts straight from the picture without first solving for a formula. That means seeing where the curve crosses each axis, where it rises or falls, and what value it approaches as x grows large in either direction. Transforming graphs means building a new function's graph out of an old one by shifting, stretching, or reflecting it, and reading the new formula back off as a set of moves rather than re-plotting from scratch. Four such moves cover every case. f(x)+k shifts the whole graph up k units. f(x-h) shifts it right h units. a\cdot f(x) stretches it vertically by a factor of a, and flips it over the x-axis if a is negative. f(-x) reflects it over the y-axis. Stacking several of these has one fixed order. Apply the inside-the-parentheses moves first: the ones acting on x itself. Then apply the outside moves last: the ones acting on the whole output. For y=-2f(x-3)+5: shift right 3 first (inside), then stretch by 2, reflect, and shift up 5 (outside). Doing the outside moves first changes which point ends up where, even though the final formula looks identical either way.
Formula
Function Transformation Rules
Recalled from College Algebra · Transformations of functions — first taught there.
Let c > 0 and a > 1. Starting from y = f(x):
Shifts
- y = f(x) + c shifts the graph up c units;
- y = f(x) - c shifts it down c units;
- y = f(x - c) shifts it right c units;
- y = f(x + c) shifts it left c units.
Reflections
- y = -f(x): reflection across the x-axis (outputs negated);
- y = f(-x): reflection across the y-axis (inputs negated).
Stretches and compressions
- y = a \cdot f(x): vertical stretch by factor a;
- y = \frac{1}{a} f(x): vertical compression by factor a;
- y = f(ax): horizontal compression by factor a;
- y = f\left(\frac{1}{a}x\right): horizontal stretch by factor a.
Changes outside f act on outputs (vertical, intuitive direction); changes inside the parentheses act on inputs (horizontal, opposite the apparent direction).
Shift it
Predict where the vertex lands once h = 2, k = -3 — then drag h and k and check. The dashed curve is f itself; it never moves, so it's what you measure the shift against.
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Stretch and reflect it
Predict what a = -2 does to the marked point on the dashed parent — then drag a through 0 and watch the tracked point flip below the axis, not just squash toward it. Try b too: it squeezes the curve sideways, and negative b flips it left-right.
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Where does the shifted parabola cross zero?
y = (x-2)^2 - 4 is f(x) = x^2 shifted right 2 and down 4. Predict the sign on each of the three intervals its roots cut the axis into — click a band to guess, and it'll tell you right away if you're right. Not sure? Open it in the graph tool and look.
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Plot the vertex
y = -2(x-3)^2 + 5 is y=x^2 shifted right 3, up 5, flipped, and stretched by 2. Place the one point where all of that lands: the vertex.
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Where this goes next
Curve sketching and finding best-value problems, several chapters ahead
A technique from three chapters ahead treats every composed function as a shift/stretch/reflect of something simpler wrapped around something else. Read g(x) = -3(x-2)^2 + 1 as "a parabola, flipped, stretched by 3, shifted right 2 and up 1" instead of just multiplying it out. That habit is what makes that technique click instead of feeling like a memorized rule.
Check your understanding (counts toward your progress)
If f(2) = 5 and g(x) = -3f(x), find g(2).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
g(2) = -3 \cdot f(2) = -3(5) = -15.
Correct answer
Also accepted:
Why
A function has a local MAX at (-1,2) and a local MIN at (5,-2), with -1<5 and no other turning points. On which interval is the function DECREASING?
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Why
Between a local max and the next local min, the function is falling -- it decreases on (-1,5), and increases on both (-\infty,-1) and (5,\infty).
Practice: Reading a transformed graph (3 problems · doesn’t affect your progress)
Why does y = f(x-3) shift the graph RIGHT, despite the minus sign looking like it should move things left?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
The minus sign does have a real effect -- it moves the graph, just in the direction opposite to what the sign alone suggests.
The sign is not just typographic decoration -- it genuinely relocates every point on the graph.
f(x-3) and f(x)-3 are different functions entirely -- one shifts horizontally, the other vertically, and they are not interchangeable.
Why
f(x-3) hits the value f(0) when x=3, three units later than the original -- every point on the graph is pushed 3 units to the right, which is exactly what "shifts right" means.
Ask what value of x makes f(x-3) match f's original output at x=0 -- work out that the input has to be 3, not -3.
Because the new graph reaches any given output 3 units later in x than the original did
How do y=f(2x) and y=f(x/2) compare, both applied to the same graph of f?
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This has the two effects backwards -- multiplying x by a number greater than 1 inside the parentheses compresses, it doesn't stretch.
They don't have the same effect on each other -- one multiplies x by 2, the other by 1/2, which are opposite moves.
They don't have the same effect on each other -- one multiplies x by 2, the other by 1/2, which are opposite moves.
Why
Inside-the-parentheses multiplication works backwards -- f(2x)'s x is scaled up before f sees it, which squeezes (compresses) the graph horizontally, while f(x/2) scales x down, stretching it.
Horizontal changes work backwards from how they look: multiplying x by a number bigger than 1 squeezes the graph, not stretches it.
f(2x) compresses the graph horizontally; f(x/2) stretches it horizontally
A graph of f crosses the x-axis only at x=2 and x=5. Find the SUM of the x-intercepts of y = f(x-1).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Shifting right 1 moves each zero 1 unit right, so the zeros of f(x-1) are at x=3 and x=6 (since x-1=2 gives x=3 and x-1=5 gives x=6). Their sum is 3+6=9.
Correct answer
Also accepted:
Why
A shift right 1 moves each x-intercept 1 unit to the right -- find each new intercept first, then add them.
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
Describe in words how g(x) = f(x + 5) - 2 shifts the graph of f.
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Inside plus means LEFT, not right — the opposite-sign rule applies here too.
The vertical part is honest, but its sign is backwards: -2 outside means down 2, not up.
You've swapped which number is inside (horizontal) and which is outside (vertical): 5 is inside, 2 is outside.
Why
x+5 inside shifts left 5 (opposite the sign), and -2 outside shifts down 2 (literal). Each point (x,y) of f moves to (x-5,\; y-2).
Handle the inside change and the outside change separately.
From College Algebra
The graph of y = (x - h)^2 has its vertex at x = 7. What is h?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
The vertex occurs where the inside is zero: x - h = 0 \Rightarrow x = h. Since the vertex is at x=7, that means h=7.
Correct answer
Also accepted:
Why
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
To graph y = -2(x + 1)^2 + 3 from y = x^2, which order of steps is correct?
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Shifting up before stretching lets the \times(-2) scale your shift: up 3 would wrongly become down 6. Stretch before the vertical shift.
The inside is x + 1, which shifts left 1 (inside changes oppose their sign). The rest of the order is right.
It matters whenever a stretch meets a vertical shift: -2 multiplies (x+1)^2 only, not the +3, so stretching must come first.
Why
Follow the formula's own order of operations on the output: form (x+1)^2 (shift left 1), multiply by -2 (stretch by 2, flip across the x-axis), then add 3 (shift up 3). The stretch must precede the vertical shift because -2 doesn't multiply the +3. Final vertex: (-1, 3), opening downward.
Apply the transformation that's innermost in the formula first — start with what's inside the parentheses.
Decide whether the stretch or the vertical shift should come first by checking exactly what the multiplier multiplies.
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
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The parabola f(x) = x^2 + 6x has its vertex (lowest point) at x = -3 (six, negative three). On which interval is f increasing?
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Walking left to right up to x=-3, you're still descending into the valley — that's the decreasing side.
The -6 is the vertex's y-value (height), not its x-value. Increasing/decreasing intervals use x-values.
That mixes up the y-value of the vertex with the x-value. The turnaround happens at the input x=-3.
Why
Walking left to right, an upward parabola goes downhill until its vertex, then uphill. The vertex is at x=-3, so f decreases on (-\infty,-3) and increases on (-3,\infty).