Sketch Graphs
Learn to sketch graphs using calculus for Victorian Year 11 Mathematical Methods (VCAA). For power functions and polynomials up to degree four, the derivative shows where a curve rises and falls.
You will learn to find intercepts and stationary points, use the first-derivative sign test to classify each turning point, locate the local and global maxima and minima, and describe the curve's behaviour for large positive and negative x.
Theory
In Year 11 Mathematical Methods (Unit 2), you sketch curves of polynomials by finding intercepts and stationary points, then using the first-derivative sign test to decide whether each is a maximum, minimum or stationary point of inflection. This page shows how to locate \(f'(x)=0\), classify the points by the sign of \(f'\), state increasing and decreasing intervals, and describe end behaviour.
A stationary point is a point where the gradient is zero, so \(f'(x)=0\). The curve has a horizontal tangent there.
The first-derivative sign test classifies each stationary point by the sign of \(f'(x)\) just before and just after it. A change from \(+\) to \(-\) is a local maximum; from \(-\) to \(+\) is a local minimum; no change of sign is a stationary point of inflection.
Where \(f'(x)>0\) the curve is increasing; where \(f'(x)<0\) it is decreasing. Together with the intercepts and the end behaviour (what happens as \(x\to\pm\infty\)), these features give the sketch.
Stationary points occur where the derivative is zero:
The nature follows from the sign of \(f'\) on each side:
How to sketch a polynomial using \(f'\)
- Intercepts: find \(f(0)\) for the \(y\)-intercept and solve \(f(x)=0\) (factorise) for the \(x\)-intercepts.
- Stationary points: differentiate, solve \(f'(x)=0\), and compute each \(y\)-coordinate from \(f(x)\).
- Classify: test the sign of \(f'\) just before and after each stationary point (max / min / inflection).
- Sketch: add the end behaviour as \(x\to\pm\infty\) and join the features smoothly.
Differentiate and solve \(f'(x)=0\):
| \(f'(x)\) | \(=\) | \(2x-6\) |
| \(2x-6\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(3\) |
\(y\)-coordinate from the curve:
| \(f(3)\) | \(=\) | \((3)^{2}-6(3)+5\) |
| \(=\) | \(-4\) |
Sign test around \(x=3\):
| \(f'(2)\) | \(=\) | \(2(2)-6=-2\ (<0)\) |
| \(f'(4)\) | \(=\) | \(2(4)-6=2\ (>0)\) |
The gradient changes \(-\to+\), so the point is a minimum.
Minimum at \((3,-4)\).
Differentiate and solve \(f'(x)=0\):
| \(f'(x)\) | \(=\) | \(3x^{2}-3\) |
| \(3x^{2}-3\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\pm 1\) |
\(y\)-coordinates:
| \(f(-1)\) | \(=\) | \((-1)^{3}-3(-1)=2\) |
| \(f(1)\) | \(=\) | \((1)^{3}-3(1)=-2\) |
Sign test (using \(f'(x)=3(x-1)(x+1)\)):
| \(f'(-2)\) | \(=\) | \(9\ (>0),\quad f'(0)=-3\ (<0)\) |
| \(f'(2)\) | \(=\) | \(9\ (>0)\) |
At \(x=-1\): \(+\to-\) (max). At \(x=1\): \(-\to+\) (min).
Maximum at \((-1,2)\); minimum at \((1,-2)\).
Intercepts — \(y\)-intercept and \(x\)-intercepts:
| \(f(0)\) | \(=\) | \(0\) |
| \(x^{2}(x-3)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(0\ \text{or}\ 3\) |
Stationary points — solve \(f'(x)=0\):
| \(f'(x)\) | \(=\) | \(3x^{2}-6x=3x(x-2)\) |
| \(x\) | \(=\) | \(0\ \text{or}\ 2\) |
\(y\)-coordinates:
| \(f(0)\) | \(=\) | \(0\) |
| \(f(2)\) | \(=\) | \((2)^{3}-3(2)^{2}=-4\) |
Sign test:
| \(f'(-1)\) | \(=\) | \(9\ (>0),\quad f'(1)=-3\ (<0)\) |
| \(f'(3)\) | \(=\) | \(9\ (>0)\) |
At \(x=0\): \(+\to-\) (max). At \(x=2\): \(-\to+\) (min). As \(x\to\infty,\ y\to\infty\); as \(x\to-\infty,\ y\to-\infty\).
Max \((0,0)\), min \((2,-4)\), \(x\)-intercepts \(0\) and \(3\).
Differentiate and solve \(f'(x)=0\):
| \(f'(x)\) | \(=\) | \(3x^{2}-6x+3\) |
| \(=\) | \(3(x-1)^{2}\) | |
| \(3(x-1)^{2}\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(1\) |
\(y\)-coordinate:
| \(f(1)\) | \(=\) | \((1)^{3}-3(1)^{2}+3(1)\) |
| \(=\) | \(1\) |
Sign test around \(x=1\):
| \(f'(0)\) | \(=\) | \(3\ (>0)\) |
| \(f'(2)\) | \(=\) | \(3\ (>0)\) |
The gradient is positive on both sides — no sign change — so it is a stationary point of inflection.
Stationary point of inflection at \((1,1)\).
Common pitfalls
Frequently asked questions
How do you find stationary points of a curve?
Differentiate, set \(f'(x)=0\) and solve for \(x\), then substitute each \(x\) back into \(f(x)\) for the \(y\)-coordinate.
How do you tell a maximum from a minimum without the second derivative?
Use the first-derivative sign test: if \(f'\) goes \(+\to-\) it is a maximum, and \(-\to+\) it is a minimum.
What is a stationary point of inflection?
A stationary point where \(f'(x)=0\) but the sign of \(f'\) does not change, so the curve keeps going the same way, as at \((1,1)\) on \(y=x^{3}-3x^{2}+3x\).
How do you find where a curve is increasing or decreasing?
Solve \(f'(x)>0\) for the increasing intervals and \(f'(x)<0\) for the decreasing intervals; the sign only changes at a stationary point.
What is end behaviour and why does it matter for a sketch?
It is what happens as \(x\to\pm\infty\). For a cubic with a positive leading term, \(y\to\infty\) as \(x\to\infty\) and \(y\to-\infty\) as \(x\to-\infty\); it fixes the overall shape.