Last Updated Sep 27, 2025

Slopes of Tangents Simplified Revision Notes for Leaving Cert Mathematics

Revision notes with simplified explanations to understand Slopes of Tangents quickly and effectively.

Learn about The Basics for your Leaving Cert Mathematics Exam. This Revision Note includes a summary of The Basics for easy recall in your Mathematics exam

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Slopes of Tangents

We've seen how to find the slope of a tangent at any given $x$ point using the first derivative :

Example

infoNote

Find the slope of the tangent at $x=\tfrac{1}{2}$ for the function $f(x)=3x^2-2x^3$.

Find the first derivative :

$$f'(x)=6x-6x^2$$

This represents the slope for any $x$, for a particular $x$, just substitute the value into $x$ :

$$f'(\tfrac{1}{2})=3(\tfrac{1}{2})-2(\tfrac{1}{2})^2=1$$

Now find the equation of the line with the slope $1$.

$$m=1$$

We also need the coordinate :

$$x=\frac{1}{2}$$$$y=3(\tfrac{1}{2})^2-2(\tfrac{1}{2})^3=\tfrac{1}{2}$$

Now find the equation of the line :

$$y-y_1=m(x-x_1)$$$$y-(\tfrac{1}{2})=1(x-\tfrac{1}{2})$$$$y=x$$

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