Question

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))?

Answer 1

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`