Question

From first principles (ie using the tangent slope method), find the slope of the following curves at the given value of x.

f(x)=2x^2−6x at x = 3

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

f(x+h)= 2(x+h)^2-6(x+h)

=2x^2+4xh+2h^2-6x-6h

lim h-->0 f(x+h)-f(x)/h

=2x^2+4xh+2h^2-6x-6h - (2x^2-6x)

=4xh+2h^2-6h/h

=h(4x+2h-6)/h

=lim h -->0 = 4x+2h-6

4x+2(0)-6
=4x-6

at x=3

f(3)=4x-6
=4(3)-6
=6

Answer 1

f(3) = 2(9) - 6(3) = 0

f(3+h) = 2(3+h)^2 - 6(3+h)
= 2(9 + 6h + h^2 - 18 - 6h
= 2h^2 + 6h

slope = lim (f(3+h) - f(3) )/h , as h--->0
= lim(2h^2 + 6h - 0)/h
= lim h(2h + 6)/h
= lim 2h + 6 , as h ---> 0
= 6

Answer 2

either way works fine. Plug in the point before or after. The first way gets you to 4x-6, and then you can use that to find the slope at any point, without having to go through all of the steps each time.