One vertical change, two different graphs
A vertical transformation changes the heights of points on a function’s graph while leaving their input values fixed. That change affects the function graph, but what happens to its tangent slopes? For example, moving every point up by the same amount changes the tangent’s height, not its steepness. Stretching the graph vertically does change its steepness, and reflecting it across the horizontal axis reverses the direction of each nonhorizontal tangent.
These observations give a direct way to predict the derivative graph. If a function is shifted vertically, its derivative graph stays the same. If the function is scaled vertically, its derivative values are scaled by the same factor. In particular, reflection across the horizontal axis changes the sign of every derivative value.
Why tangent slopes transform this way
A tangent slope is the limit of the slopes of nearby secant lines. For a function \(f\), the secant slope from input \(x\) to input \(x+h\), where \(h\ne 0\), is \(\frac{f(x+h)-f(x)}{h}\). If \(f\) is differentiable at \(x\), these secant slopes approach \(f'(x)\) as \(h\) approaches zero.
Now transform the graph vertically by setting \(g(x)=a f(x)+c\). The output difference between the two transformed points is \(a\) times the original output difference: the added \(c\) cancels. The horizontal difference remains \(h\). Therefore, each transformed secant slope is \(a\) times the original secant slope. Taking the limit proves the derivative relationship.
If \(f'(x)\) exists, the limit of the right-hand side as \(h\to 0\) is \(a f'(x)\). Thus \(g'(x)=a f'(x)\). This argument also explains the geometry: vertical differences are multiplied by \(a\), while horizontal differences are unchanged, so tangent slopes are multiplied by \(a\).
The final condition matters. When \(a\ne 0\), the equation \(f(x)=(g(x)-c)/a\) expresses \(f\) as a vertical rescaling and shift of \(g\), so differentiability transfers in both directions. When \(a=0\), however, \(g(x)=c\) regardless of the behavior of \(f\). A constant function is differentiable even if the original function is not.
Reading the transformation on a derivative graph
The derivative graph plots input \(x\) against derivative value \(f'(x)\). The theorem says that the derivative value of \(g\) at each input is \(a f'(x)\). Therefore, each point \((x,f'(x))\) on the derivative graph moves to \((x,a f'(x))\). The input coordinate does not change, and the constant vertical shift \(c\) of the original function does not shift the derivative graph.
| Change to the function | Change to its derivative values |
|---|---|
| Add a constant: \(g(x)=f(x)+c\) | \(g'(x)=f'(x)\); the derivative graph is unchanged |
| Multiply outputs by a positive factor: \(g(x)=a f(x)\), \(a>0\) | \(g'(x)=a f'(x)\); derivative heights are scaled by \(a\) |
| Reflect outputs: \(g(x)=-f(x)\) | \(g'(x)=-f'(x)\); derivative signs are reversed |
| Combine scaling and shifting: \(g(x)=a f(x)+c\) | \(g'(x)=a f'(x)\); the shift has no effect |
If \(a\) is negative, the derivative graph is reflected across its horizontal axis and its vertical distances from that axis are multiplied by \(|a|\). If \(0<a<1\), the derivative graph is compressed toward the horizontal axis. If \(a>1\), it is stretched away from the axis. A zero derivative value remains zero under every scaling.
Worked examples
Worked Example: Shifting a parabola upward
Let \(f(x)=x^2\), and shift its graph upward by 3 units to form \(g(x)=x^2+3\). Find and compare the tangent slopes at \(x=2\), then compare the tangent lines.
First calculate the derivative value for \(f\) from the difference quotient. We have \(f(2)=4\) and \(f(2+h)=(2+h)^2=4+4h+h^2\), so
As \(h\to 0\), this quotient approaches \(4\), so \(f'(2)=4\). For the shifted function, \(g(2)=7\) and \(g(2+h)=7+4h+h^2\). Thus
Therefore \(g'(2)=4\), the same slope as before. The tangent to \(f\) at \((2,4)\) is \(y-4=4(x-2)\); the tangent to \(g\) at \((2,7)\) is \(y-7=4(x-2)\). Both have slope \(4\), and the second tangent is 3 units higher. The derivative check is \(g'(x)=2x\), so \(g'(2)=4\).
Worked Example: Scaling and reflecting a graph
Let \(f(x)=x^3-2x\), and define \(G(x)=-2f(x)+5\). Find the tangent slope to each graph at \(x=1\), and verify the transformed tangent slope directly.
For \(f\), \(f(1)=1-2=-1\), and
The limit as \(h\to 0\) is \(1\), so \(f'(1)=1\). The transformation has \(a=-2\), so the theorem predicts \(G'(1)=(-2)(1)=-2\). Also, \(G(1)=-2(-1)+5=7\). Directly, \(G(1+h)=-2f(1+h)+5=7-2h-6h^2-2h^3\), and therefore
Thus \(G'(1)=-2\), as predicted. The tangent to \(G\) at \((1,7)\) is \(y-7=-2(x-1)\). The negative scale reflects the graph vertically and reverses the original positive tangent slope. As a further check, \(G(x)=-2x^3+4x+5\), whose derivative is \(-6x^2+4\); at \(x=1\), this gives \(-6+4=-2\).
Worked Example: Transforming selected derivative-graph values
Suppose \(f\) is differentiable at inputs \(-1\), \(0\), and \(2\), and its derivative graph gives \(f'(-1)=-4\), \(f'(0)=0\), and \(f'(2)=6\). Define \(q(x)=-\frac{1}{2}f(x)+3\). Find the corresponding values on the derivative graph of \(q\).
Here \(a=-\frac{1}{2}\) and \(c=3\), so \(q'(x)=-\frac{1}{2}f'(x)\). At each input:
The transformed derivative-graph points are therefore \((-1,2)\), \((0,0)\), and \((2,-3)\). The shift by 3 in the function contributes nothing to these derivative values. The negative factor reverses the nonzero signs, and the factor of \(\frac{1}{2}\) halves their magnitudes. Each result checks against \(q'(x)=-\frac{1}{2}f'(x)\) at its input.
A common mistake: shifting the derivative graph too
A frequent error is to carry the vertical shift of the function into its derivative. For example, a learner might write
This adds the constant 5 to the derivative, as though the derivative graph had been shifted upward by 5. But the constant changes the function’s height, not its tangent slopes. Using the difference quotient makes the cancellation explicit:
So the correct derivative is \(2x\), not \(2x+5\). A reliable check is to ask which quantities change between nearby points: a constant shift changes neither their output difference nor their horizontal difference, and therefore cannot change the secant slope.
When the result is useful
Vertical transformations let you compare derivative graphs without recomputing each tangent slope from scratch. If a known function is shifted, scaled, or reflected, the derivative relationship follows from how the graph’s vertical differences change. This is especially useful when a graph or table supplies derivative values at selected inputs: multiply each value by the scale factor, and do not add the function’s vertical shift.
Keep track of which graph is being transformed. For \(g(x)=a f(x)+c\), the original function graph moves by the rule \(y\mapsto ay+c\), but the derivative graph changes by \(y\mapsto ay\). The \(+c\) belongs to the function’s output heights, not to its tangent slopes. If \(a<0\), retain the sign: a negative derivative becomes positive and a positive derivative becomes negative, while a zero derivative stays zero.
Write it as \(g(x)=a f(x)+c\), if possible.
The rule applies wherever \(f\) is differentiable; if \(a=0\), the transformed function is constant.
Use \(g'(x)=a f'(x)\). Do not include \(c\).
Inputs stay fixed, derivative heights scale by \(a\), and a negative \(a\) reverses signs.
Check Your Understanding
Use tangent slopes and the vertical-transformation rule to answer the following.
- If \(g(x)=f(x)+7\) and \(f'(3)=-2\), what is \(g'(3)\), and why?
- If \(h(x)=4f(x)\) and \(f'(x)=\frac{3}{2}\) at an input, what is \(h'(x)\) there?
- If \(p(x)=-f(x)\), how does each point on the derivative graph of \(f\) move?
- For \(r(x)=-3f(x)+2\), state the relationship between \(r'(x)\) and \(f'(x)\). Which constant has no effect on the derivative?
- Why does the theorem’s differentiability conclusion need special care when the scale factor is zero?
