Gradients, Tangents and Normals

Example 1: Finding the Tangent

A tangent is a straight line which touches our graph, but doesn’t pass through it at the meeting point. By definition, a straight line graph (i.e. $y = x$) cannot have a tangent – only a curved graph can have a tangent.

So, as an example, here’s the graph of $textcolor{blue}{y = (x – 1)^2 – 2}$. Find the equation of the tangent line at $x = dfrac{3}{2}$.

[4 marks]

To find the tangent, we first find the gradient at our point of interest.

Since we have

$textcolor{blue}{y = (x – 1)^2 – 2}$

$= textcolor{blue}{x^2 – 2x – 1}$

we have a gradient of

$dfrac{dy}{dx} = 2x – 2$

at any point, and $dfrac{dy}{dx} = 1$ when $x = dfrac{3}{2}$.

Since our gradient is a straight line, it must have the general equation

$y = mx + c$

When $x = dfrac{3}{2}$, $y = -dfrac{7}{4}$ and $dfrac{dy}{dx} = m_T = 1$.

Then we can plug in these values to find $c$:

$-dfrac{7}{4} = dfrac{3}{2} + c$

$c = -dfrac{7}{4} – dfrac{3}{2} = -dfrac{13}{4}$

Therefore, our tangent to $y = (x – 1)^2 – 2$ at $x = dfrac{3}{2}$ is given by the equation $textcolor{red}{y = x – dfrac{13}{4}}$.

Example 2: Finding the Normal

The normal is a straight line which is exactly perpendicular to the tangent.

Say we have the same graph from Example 1, but now we wish to find the normal at the point $x = dfrac{3}{2}$.

[3 marks]

Since we already know that the gradient of the tangent, $m_T = 1$, we can conclude that the gradient of the normal, $m_N = -1$.

Given, also, that the point of interest is at $left( dfrac{3}{2}, -dfrac{7}{4} right)$, we can form a straight line equation for the normal:

$-dfrac{7}{4} = left(-1 times dfrac{3}{2} right) + c$

$c = -dfrac{7}{4} + dfrac{3}{2} = -dfrac{1}{4}$

The equation of the normal is given by $textcolor{limegreen}{y = -x – dfrac{1}{4}}$.