*f*(*x*) =
2*x*^{2} + 3*x* so

`f(x+h)=2(x+h)^2+3(x+h)`

`=2(x^2+2xh+h^2)+` `(3x+3h)`

`=2x^2+4xh+2h^2+` `3x+` `3h`

We now need to find:

`(dy)/(dx)=lim_(h->0)(f(x+h)-f(x))/h`

`=lim_(h->0)([2x^2+4xh+2h^2+3x+3h]-[2x^2+3x])/h`

`=lim_(h->0)(4xh+2h^2+3h)/h`

`=lim_(h->0)(4x+2h+3)`

`=4x+3`

We have found an expression that can give us the slope of the tangent anywhere on the curve.

If `x = -2`, the slope is `4(-2) + 3 = -5` (red, in the graph below)

If `x = 1`, the slope is `4(1) + 3 = 7` (green)

If `x = 4`, the slope is `4(4) + 3 = 19` (black)

We can see that our answers are correct when we graph the curve (which is a parabola) and observe the slopes of the tangents.

This is what makes calculus so powerful. We can find the slope anywhere on the curve (i.e. the rate of change of the function anywhere).

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