Why do you need both f(a) and f prime of a?

f prime of a gives the gradient of the tangent, and f(a) gives the y-coordinate of the point the line must pass through. You need both to write the line, because point-slope form uses the point (a, f(a)) as well as the gradient.

Year 12 Methods (Unit 3 & 4) Further differentiation and applications

Tangents and normals

Q: How do you find the equation of a tangent to a curve?

Find the point by evaluating f(a), find the gradient by evaluating the derivative f prime of a, then substitute into y minus f(a) equals f prime of a times (x minus a). For y equals x squared at x equals 3, the tangent is y equals 6x minus 9.

Q: What is the gradient of the normal to a curve?

The negative reciprocal of the tangent gradient, minus 1 divided by f prime of a. If the tangent gradient is 4, the normal gradient is minus one quarter.

Q: What is the difference between a tangent and a normal?

The tangent touches the curve with gradient f prime of a; the normal is perpendicular to it at the same point with gradient minus 1 divided by f prime of a.

Q: How do you find where the tangent is horizontal?

Solve f prime of x equals 0. For y equals x cubed minus 3x this gives x equals minus 1 and x equals 1, with horizontal tangents y equals 2 and y equals minus 2.

Q: Why do you need both f(a) and f prime of a?

f prime of a is the gradient and f(a) is the y-coordinate of the point the line passes through; point-slope form needs both.

Q: How do you find where a tangent meets the x-axis?

Find the tangent equation, set y equal to 0 and solve for x. For y equals 6x minus 9 this gives x equals 1.5.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), the tangent to \(y=f(x)\) at \(x=a\) is the line touching the curve there, with gradient \(f'(a)\); the normal is perpendicular to it, with gradient \(-\dfrac{1}{f'(a)}\). Find the point \((a,f(a))\) and the gradient, then use point–slope form — applied to polynomial, exponential, logarithmic and trigonometric curves, to horizontal tangents, and to where a line meets an axis.

The tangent to a curve \(y=f(x)\) at the point where \(x=a\) is the straight line that just touches the curve there and has the same gradient as the curve. That gradient is the derivative evaluated at the point, \(f'(a)\).

The normal at the same point is the straight line perpendicular to the tangent. Perpendicular lines have gradients whose product is \(-1\), so the normal's gradient is the negative reciprocal \(-\dfrac{1}{f'(a)}\).

To write either line you need two things: the point \((a,f(a))\) — found by substituting \(x=a\) into \(f\) — and the gradient. Then substitute into the point–slope form \(y-f(a)=m(x-a)\).

Key idea. Tangent gradient \(=f'(a)\); normal gradient \(=-\dfrac{1}{f'(a)}\). Both lines pass through \((a,f(a))\), so always compute \(f(a)\) as well as \(f'(a)\).

Tangent (gradient \(f'(1)=2\)) and normal (gradient \(-\tfrac{1}{2}\)) to \(y=x^{2}\) at \((1,1)\)

Horizontal tangents where \(f'(x)=0\): \(y=2\) at \((-1,2)\) and \(y=-2\) at \((1,-2)\)

Tangent to \(y=f(x)\) at \(x=a\):

\[y-f(a)=f'(a)\,(x-a)\]

y - f (a) = f^{'} (a) (x - a)

Normal to \(y=f(x)\) at \(x=a\) (provided \(f'(a)\neq 0\)):

\[y-f(a)=-\dfrac{1}{f'(a)}\,(x-a)\]

y - f (a) = - \frac{1}{f^{'} (a)} (x - a)

Perpendicular gradients (why the normal uses the negative reciprocal):

\[m_{\text{tangent}}\times m_{\text{normal}}=-1\]

Horizontal tangent. The tangent is horizontal where \(f'(x)=0\); there the tangent is \(y=f(a)\) and the normal is the vertical line \(x=a\). To find where a tangent cuts an axis, set \(y=0\) (for the \(x\)-intercept) or \(x=0\) (for the \(y\)-intercept) in the tangent equation.

How to find a tangent or normal at \(x=a\)

Find the point. Evaluate \(f(a)\) to get \((a,f(a))\).
Differentiate. Find \(f'(x)\), then evaluate \(f'(a)\) — this is the tangent gradient.
Choose the gradient. For the tangent use \(m=f'(a)\); for the normal use \(m=-\dfrac{1}{f'(a)}\).
Write the line. Substitute into \(y-f(a)=m(x-a)\) and simplify.
Answer the extra part. For a horizontal tangent solve \(f'(x)=0\); for an axis intercept set \(y=0\) or \(x=0\) in the line.

Transcendental curves. The same method works for \(y=e^{x}\), \(y=\ln x\) and \(y=\sin x,\cos x,\tan x\) — only the derivative changes (e.g. \(\dfrac{d}{dx}e^{x}=e^{x}\), \(\dfrac{d}{dx}\ln x=\dfrac{1}{x}\), \(\dfrac{d}{dx}\cos x=-\sin x\)).

Example 1 — Tangent to a polynomial

Find the tangent to \(y=x^{2}\) at \(x=3\).

Solution

\(f(3)=9\); \(f'(x)=2x\) so \(f'(3)=6\).

\(y-9\)	\(=\)	\(6(x-3)\)
\(y\)	\(=\)	\(6x-9\)

y = 6 x - 9

Example 2 — Normal to a polynomial

Find the normal to \(y=x^{2}\) at \(x=3\).

Solution

\(f(3)=9\), \(f'(3)=6\), so \(m=-\dfrac{1}{6}\).

\(y-9\)	\(=\)	\(-\dfrac{1}{6}(x-3)\)
\(y\)	\(=\)	\(-\dfrac{1}{6}x+\dfrac{19}{2}\)

y = - \frac{1}{6} x + \frac{19}{2}

Example 3 — Tangent to \(y=e^{x}\)

Find the tangent to \(y=e^{x}\) at \(x=0\).

Solution

\(f(0)=1\); \(f'(x)=e^{x}\) so \(f'(0)=1\).

\(y-1\)	\(=\)	\(1(x-0)\)
\(y\)	\(=\)	\(x+1\)

y = x + 1

Example 4 — Horizontal tangent

Where is the tangent to \(y=x^{3}-3x\) horizontal?

Solution

A horizontal tangent has gradient \(0\), so solve \(f'(x)=0\).

\(3x^{2}-3\)	\(=\)	\(0\)
\(x\)	\(=\)	\(-1\) or \(1\)

Points \((-1,2)\) and \((1,-2)\); tangents \(y=2\) and \(y=-2\).

x = - 1 or x = 1

Common pitfalls

The normal gradient is the negative reciprocal. Use \(m=-\dfrac{1}{f'(a)}\), not \(f'(a)\) and not \(-f'(a)\). If \(f'(a)=4\) the normal gradient is \(-\dfrac{1}{4}\).

Compute \(f(a)\), not just \(f'(a)\). The gradient alone is not enough — the line must pass through \((a,f(a))\), so substitute the point to get the constant term.

Watch the point–slope algebra. Expand \(y-f(a)=m(x-a)\) carefully; a dropped \(f(a)\) or a sign slip changes the intercept and the whole line.

Frequently asked questions

How do you find the equation of a tangent to a curve?

Find \(f(a)\) for the point, \(f'(a)\) for the gradient, then substitute into \(y-f(a)=f'(a)(x-a)\). For \(y=x^{2}\) at \(x=3\) this is \(y=6x-9\).

What is the gradient of the normal to a curve?

The negative reciprocal of the tangent gradient, \(-\dfrac{1}{f'(a)}\). If \(f'(a)=4\), the normal gradient is \(-\dfrac{1}{4}\).

What is the difference between a tangent and a normal?

The tangent touches the curve with gradient \(f'(a)\); the normal is perpendicular to it at the same point, with gradient \(-\dfrac{1}{f'(a)}\).

How do you find where the tangent is horizontal?

Solve \(f'(x)=0\). For \(y=x^{3}-3x\), \(f'(x)=3x^{2}-3=0\) gives \(x=\pm 1\), with tangents \(y=2\) and \(y=-2\).

Why do you need both f(a) and f'(a)?

\(f'(a)\) is the gradient and \(f(a)\) is the \(y\)-coordinate of the point the line passes through; point–slope form needs both.

How do you find where a tangent meets the x-axis?

Find the tangent equation, set \(y=0\) and solve for \(x\). For \(y=6x-9\), \(y=0\) gives \(x=1.5\).

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