Find the tangent to the curve y = x^2 + 3x + 2 at x = 1

First, we differentiate y = x^2 + 3x + 2 to find the slope of the curve at any given x point:

dy/dx = 2x + 3

So at x = 1, dy/dx = 2 + 3 = 5

We know that straight lines come in the form y = mx + c, and in this case m is equal to our gradient which we worked out earlier - 5.

So y = 5x + c, and we want to solve for c. Luckily we know a way to find an x and y co-ordinate that lie on the curve - the original function!

At x = 1, y = 1^2+3*1+2, so y = 6, and the point (1,6) lies on the curve.

Using this, we can solve y = 5x + c by substitution:

6 = 5*1 + c

1 = c

Hence the equation of the tangent is y = 5x + 1.