Tangents And Normals
Master tangents and normals for Queensland Year 11 Mathematical Methods (QCAA). The tangent is the straight line touching a curve at a point, and the normal is the line at right angles to it there.
You will learn to find a curve's gradient with the derivative, write the equation of the tangent, and use the negative reciprocal to write the equation of the normal — a skill applied across geometry and physics.
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 \(x=a\) is the line that just touches it, with gradient \(f'(a)\); the normal is the perpendicular line through the same point, with gradient \(-\dfrac{1}{f'(a)}\). This page shows how to find the equation of a tangent and a normal, horizontal tangents, and where they cut the axes.
The tangent to \(y=f(x)\) at the point \((a,\,f(a))\) is the straight line that touches the curve there and has the same steepness as the curve. Its gradient is the value of the derivative at that point, \(m_T=f'(a)\).
The normal at the same point is the straight line through \((a,\,f(a))\) that is perpendicular to the tangent. Perpendicular gradients multiply to \(-1\), so the normal gradient is the negative reciprocal \(m_N=-\dfrac{1}{f'(a)}\).
At the point where \(x=a\), with gradient \(f'(a)\):
How to find a tangent or normal at \(x=a\)
- Point: substitute \(a\) into the curve to get \((a,\,f(a))\).
- Gradient: differentiate to get \(f'(x)\), then evaluate \(f'(a)\). For a normal, take \(-\dfrac{1}{f'(a)}\).
- Line: substitute the point and gradient into \(y-f(a)=m(x-a)\) and simplify.
Point — substitute \(x=3\) into the curve:
| \(f(3)\) | \(=\) | \((3)^2-2(3)\) |
| \(=\) | \(9-6\) | |
| \(=\) | \(3\) |
So the point of contact is \((3,\,3)\).
Gradient — differentiate, then substitute:
| \(f'(x)\) | \(=\) | \(2x-2\) |
| \(m_T=f'(3)\) | \(=\) | \(2(3)-2\) |
| \(=\) | \(4\) |
Line — use \(y-y_1=m(x-x_1)\):
| \(y-3\) | \(=\) | \(4(x-3)\) |
| \(y-3\) | \(=\) | \(4x-12\) |
| \(y\) | \(=\) | \(4x-12+3\) |
| \(y\) | \(=\) | \(4x-9\) |
Tangent: \(y=4x-9\).
Point:
| \(f(1)\) | \(=\) | \((1)^2-4(1)+5\) |
| \(=\) | \(1-4+5\) | |
| \(=\) | \(2\) |
The point is \((1,\,2)\).
Tangent gradient:
| \(f'(x)\) | \(=\) | \(2x-4\) |
| \(f'(1)\) | \(=\) | \(2(1)-4\) |
| \(=\) | \(-2\) |
Normal gradient (negative reciprocal):
| \(m_N\) | \(=\) | \(-\dfrac{1}{f'(1)}=-\dfrac{1}{-2}\) |
| \(=\) | \(\dfrac{1}{2}\) |
Line — then clear the fraction:
| \(y-2\) | \(=\) | \(\tfrac12(x-1)\) |
| \(2(y-2)\) | \(=\) | \(x-1\) |
| \(2y-4\) | \(=\) | \(x-1\) |
| \(x-2y+3\) | \(=\) | \(0\) |
Normal: \(x-2y+3=0\).
Horizontal means gradient \(0\), so solve \(f'(x)=0\):
| \(f'(x)\) | \(=\) | \(2x-6\) |
| \(2x-6\) | \(=\) | \(0\) |
| \(2x\) | \(=\) | \(6\) |
| \(x\) | \(=\) | \(3\) |
\(y\)-coordinate — substitute into the curve:
| \(f(3)\) | \(=\) | \((3)^2-6(3)+5\) |
| \(=\) | \(9-18+5\) | |
| \(=\) | \(-4\) |
Horizontal tangent at \((3,\,-4)\).
Point and gradient:
| \(f(2)\) | \(=\) | \((2)^2-3=1\) |
| \(f'(x)\) | \(=\) | \(2x\) |
| \(f'(2)\) | \(=\) | \(2(2)=4\) |
Tangent line:
| \(y-1\) | \(=\) | \(4(x-2)\) |
| \(y-1\) | \(=\) | \(4x-8\) |
| \(y\) | \(=\) | \(4x-7\) |
Set \(y=0\) for the \(x\)-axis:
| \(0\) | \(=\) | \(4x-7\) |
| \(4x\) | \(=\) | \(7\) |
| \(x\) | \(=\) | \(\tfrac{7}{4}\) |
Meets the \(x\)-axis at \(\left(\tfrac{7}{4},\,0\right)\).
Common pitfalls
Frequently asked questions
How do you find the equation of a tangent to a curve?
Differentiate to get \(f'(x)\), evaluate \(f'(a)\) for the gradient, find the point \((a,f(a))\) from the curve, then use \(y-f(a)=f'(a)(x-a)\).
What is the gradient of the normal to a curve?
The negative reciprocal of the tangent gradient, \(-\dfrac{1}{f'(a)}\), because the normal is perpendicular to the tangent.
What is the difference between a tangent and a normal?
The tangent just touches the curve and has gradient \(f'(a)\); the normal goes through the same point but at right angles to the tangent.
How do you find where a curve has a horizontal tangent?
Set \(f'(x)=0\), solve for \(x\), then substitute back into \(f(x)\) for the \(y\)-coordinate.
Do you use the derivative to find the y coordinate of the point?
No — the \(y\)-coordinate comes from the curve \(f(a)\). The derivative only gives the gradient.