Sketch Graphs
Learn to sketch graphs using calculus for Queensland Year 11 Mathematical Methods (QCAA). 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.
Every question with a fully worked solution.
- Sketch Graphs - Video - Curve Sketching with Asymptotes Watch
Theory
In Year 11 Mathematical Methods (QCAA, 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.