# How do you find the slope of the secant lines of  f (x) = x^2 − x − 42 at (−5, −12), and (7, 0)?

Oct 30, 2015

Assuming that the word "lines" should be "line", it is exactly the same as the slope of the line through the points (−5, −12), and (7, 0).

#### Explanation:

If my interpretation is incorrect, then perhaps you asking for the general equation of a secant line to the curve that includes the point $\left(- 5 , - 12\right)$

If that is the question, then the form of the answer will depend on whether you call the second point $\left(x , f \left(x\right)\right)$ or $\left(a , f \left(a\right)\right)$ or somthing similar of call it $\left(- 5 + h , f \left(- 5 + h\right)\right)$

For $\left(x , f \left(x\right)\right)$, we find slope:
m = (f(x)-(-12))/(x-(-5)) = (x^2-x-42 + 12)/(x+5) = (x^2-x-30)/x+5)

$= \frac{\left(x - 6\right) \left(x + 5\right)}{x + 5} = x - 6$ (for $x \ne - 5$)

For $\left(- 5 + h , f \left(- 5 + h\right)\right)$, we find slope:

$m = \frac{f \left(- 5 + h\right) - \left(- 12\right)}{\left(- 5 + h\right) - \left(- 5\right)}$

$= \frac{{\left(- 5 + h\right)}^{2} - \left(- 5 + h\right) - 42 + 12}{- 5 + h + 5}$

$= \frac{25 - 10 h + {h}^{2} + 5 - h - 42 + 12}{h}$

$= \frac{- 10 h + {h}^{2} - h}{h} = - 11 + h$ (for $h \ne 0$)

The general equation of a secant line to the curve that includes the point $\left(7 , 0\right)$ is found by similar methods.