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
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