a. Note: y means "the first derivative". This can also be written `dy/dx`.

Now f(x) = x2 + 4x

So

`f(x+h)=(x+h)^2+4(x+h)`

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

Therefore

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

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

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

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

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

`=2x+4`

b. When `x = 1`, slope `m = 2(1) + 4 = 6`

When `x = -6`, slope `m = 2(-6) + 4 = -8`

c. Sketch:

curve showing different slopes