How do you find the slope of the secant lines of # f(x) = x^2 + 5x# at (6 , f(6)) and (6 + h , f(6 + h))?

1 Answer
Steve M
Jan 23, 2017

Slope #= h+17#

Explanation:

We have # f(x)=x^2+5x #

When #x=6#:

# f(x) = 6^2+5*6 #
# \ \ \ \ \ \ \ = 36 + 30 #
# \ \ \ \ \ \ \ = 66 #

When #x=6+h#:

# f(x) = (6+h)^2+5(6+h) #
# \ \ \ \ \ \ \ = 36 + 12h+h^2+30+5h #
# \ \ \ \ \ \ \ = h^2+17h+66 #

So the slope of the secant line at #(6,f(6))# and #(6+h, f(6+h))# is;

# (Delta y)/(Delta x) = (f(6+h)-f(6))/((6+h)-6) #
# \ \ \ \ \ \ \ = (h^2+17h+66 -66) / (h) #
# \ \ \ \ \ \ \ = (h^2+17h) / (h) #
# \ \ \ \ \ \ \ = h+17 #

Side Note - What is the significance of this:
If we let #h rarr 0# then the secant line becomes the tangent line, and that tangent would therefore have slope #17# when #x=6#.

With our knowledge of Calculus we can confirm this as:

# f'(x)=2x+5 => f'(6)=12+5 = 17 #

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