Finding the gradient of a curve

​​In a nutshell

Curves don't have a constant gradient, but using tangents can allow you to estimate the gradient of a curve at a particular point on it. This is because the gradient of the curve at some point $$A$$ is equal to the gradient of the tangent at point $$A$$. The gradient of the tangent would be an estimate of the gradient of the curve, since it can be difficult to sketch a perfectly accurate tangent at a point on a curve.

Gradient of a tangent

A tangent is a straight line with equation $$y=mx+c$$. The gradient of a tangent can be calculated using: 

​$$\boxed{m=\dfrac{\text{change in }y}{\text{change in }x}}$$​​

Thus, when a tangent is drawn at a point on a curve, the coordinates of two points on that line must be identified and used.

Note: Consider a tangent at a turning point. It would be a horizontal line and thus has a zero gradient. Therefore, it follows that the turning points of curves have gradient of zero.

Example 1

Consider the curve given by the equation $$y=x^2$$. Below, this curve is pictured with a tangent at the point $$(2,4)$$:

Find the gradient of the curve at the point $$(2,4)$$.

You need to calculate the gradient of the pictured tangent. You already know one point on the curve: $$(2,4)$$, but you need another point to calculate the gradient. The point $$(4,12)$$ is also on this line:

Hence, you can now calculate the gradient using $$\frac{\text{change in }y}{\text{change in }x}$$: 

​$$m=\dfrac{12-4}{4-2}=\dfrac{8}{2}=4$$​​

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Therefore, the gradient of $$y=x^2$$ at the point $$(2,4)$$ is $$\underline{4}$$.

Example 2

Below, the graph given by the equation $$y=x^3+2x^2-x$$ is presented, with a tangent at the point $$(-1,2)$$:

Note: This tangent does cross the curve, but not at the point $$(-1,2)$$. It only touches the curve there. So it is still a tangent, but specifically it is a tangent at $$(-1,2)$$ and not at $$(0,0)$$, where it crosses through the curve.

Using this diagram, calculate the gradient of the curve $$y=x^3+2x^2-x$$ at the point $$(-1,2)$$.

You need a second point on the line to calculate its gradient. The point $$(0,0)$$ is also on the line. 

Note: It is incidental and unimportant that $$(0,0)$$ is also on the curve itself. Here you are using two points on a line to find the gradient of said line.

The gradient of the tangent is:

$$m=\dfrac{0-2}{0-(-1)}=\dfrac{-2}1=-2$$​​

Therefore, the gradient of $$y=x^3+2x^2-x$$ at the point $$(-1,2)$$ is $$\underline{-2}$$.

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