The Tangent To A Curve At A Point
Understand the tangent to a curve at a point for Queensland Year 11 Mathematical Methods (QCAA). A tangent just touches a curve at one point, and its steepness is the limit of the gradients of secant lines as the second point slides closer.
You will learn to build secant gradients, watch them approach the tangent gradient, and find it exactly from first principles — the foundation of differential calculus.
Every question with a fully worked solution.
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Theory
In Year 11 Mathematical Methods (QCAA, Unit 2), the tangent to a curve \(y=f(x)\) at a point \(P\) is the straight line that just touches the curve there. Its gradient is the limit of the gradients of secants (chords) through \(P\) as the second point slides in to \(P\). This page shows how to find that gradient from first principles and how the tangent gradient is the instantaneous rate of change.
A secant (chord) joins two points on the curve. Its gradient is the average rate of change, \(\dfrac{\Delta y}{\Delta x}\), between those two points.
The tangent at \(P\) is the limiting position of the secant through \(P\) as the second point \(Q\) approaches \(P\). Its gradient is the derivative at that point, written \(f'(a)\), and equals the instantaneous rate of change there.
Using first principles, the gradient of the tangent at \(x=a\) is \(\displaystyle f'(a)=\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}\): the gradient of the chord from \((a,f(a))\) to \((a+h,f(a+h))\), taken as \(h\to0\).
Gradient of the secant (average rate of change) from \(x=a\) to \(x=a+h\):
Gradient of the tangent at \(x=a\) (first principles):
Finding the tangent gradient at \(x=a\) from first principles
- Set up the chord gradient \(\dfrac{f(a+h)-f(a)}{h}\): substitute \(a+h\) and \(a\) into \(f\).
- Expand \(f(a+h)\) and subtract \(f(a)\), then simplify the numerator.
- Cancel the common factor \(h\) so no \(h\) is left in the denominator.
- Take the limit as \(h\to0\): the remaining expression is the tangent gradient \(f'(a)\).
Endpoints — substitute into the curve:
| \(f(1)\) | \(=\) | \((1)^2=1\) |
| \(f(4)\) | \(=\) | \((4)^2=16\) |
So the chord joins \((1,\,1)\) and \((4,\,16)\).
Gradient — average rate of change \(\dfrac{\Delta y}{\Delta x}\):
| \(m_{\text{sec}}\) | \(=\) | \(\dfrac{16-1}{4-1}\) |
| \(=\) | \(\dfrac{15}{3}\) | |
| \(=\) | \(5\) |
Gradient of the chord: \(5\).
Chord gradient — substitute \(x=2\) and \(x=2+h\):
| \(m_{\text{sec}}\) | \(=\) | \(\dfrac{(2+h)^2-(2)^2}{h}\) |
| \(=\) | \(\dfrac{4+4h+h^2-4}{h}\) | |
| \(=\) | \(\dfrac{4h+h^2}{h}\) |
Cancel \(h\):
| \(m_{\text{sec}}\) | \(=\) | \(\dfrac{h(4+h)}{h}\) |
| \(=\) | \(4+h\) |
As \(h\to0\) the chords give \(4.1,\,4.01,\,4.001,\dots\)
Take the limit \(h\to0\):
| \(f'(2)\) | \(=\) | \(\lim_{h\to0}\,(4+h)\) |
| \(=\) | \(4\) |
Tangent gradient at \(x=2\) is \(4\).
Values — substitute into \(f(x)=x^2-3x\):
| \(f(4)\) | \(=\) | \((4)^2-3(4)=4\) |
| \(f(4+h)\) | \(=\) | \((4+h)^2-3(4+h)\) |
| \(=\) | \(16+8h+h^2-12-3h\) | |
| \(=\) | \(4+5h+h^2\) |
Chord gradient, then cancel \(h\):
| \(\dfrac{f(4+h)-f(4)}{h}\) | \(=\) | \(\dfrac{(4+5h+h^2)-4}{h}\) |
| \(=\) | \(\dfrac{5h+h^2}{h}\) | |
| \(=\) | \(5+h\) |
Take the limit \(h\to0\):
| \(f'(4)\) | \(=\) | \(\lim_{h\to0}\,(5+h)\) |
| \(=\) | \(5\) |
Tangent gradient at \(x=4\) is \(5\).
Expand \(f(x+h)-f(x)\):
| \(f(x+h)\) | \(=\) | \((x+h)^2-(x+h)\) |
| \(=\) | \(x^2+2xh+h^2-x-h\) | |
| \(f(x+h)-f(x)\) | \(=\) | \(2xh+h^2-h\) |
Chord gradient, cancel \(h\), take the limit:
| \(\dfrac{f(x+h)-f(x)}{h}\) | \(=\) | \(\dfrac{h(2x+h-1)}{h}\) |
| \(=\) | \(2x+h-1\) | |
| \(f'(x)\) | \(=\) | \(\lim_{h\to0}(2x+h-1)=2x-1\) |
Point and gradient at \(x=3\):
| \(f(3)\) | \(=\) | \((3)^2-3=6\) |
| \(f'(3)\) | \(=\) | \(2(3)-1=5\) |
Tangent line \(y-y_1=m(x-x_1)\):
| \(y-6\) | \(=\) | \(5(x-3)\) |
| \(y\) | \(=\) | \(5x-15+6\) |
| \(y\) | \(=\) | \(5x-9\) |
\(f'(x)=2x-1\); tangent at \(x=3\) is \(y=5x-9\).
Common pitfalls
Frequently asked questions
What is the tangent to a curve at a point?
The straight line that just touches the curve at that point and has the same steepness as the curve there. Its gradient is \(f'(a)\).
How is a secant different from a tangent?
A secant (chord) cuts the curve at two points and gives an average rate of change; the tangent is the limit of those secants as the second point slides in to the first.
How do you find the tangent gradient from first principles?
Form the chord gradient \(\dfrac{f(a+h)-f(a)}{h}\), simplify, cancel the factor \(h\), then take the limit as \(h\to0\).
Why do you cancel \(h\) before taking the limit?
Substituting \(h=0\) straight away gives \(\dfrac00\), which is undefined. Cancelling the common factor \(h\) removes it from the denominator so the limit can be read off.
Is the tangent gradient the same as the instantaneous rate of change?
Yes. The tangent gradient \(f'(a)\) is the instantaneous rate of change of \(y\) with respect to \(x\) at \(x=a\).