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

Question

Cameron Alexander · Accepted Answer

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).

If my interpretation is incorrect, then perhaps you asking for the general equation of a secant line to the curve that includes the point #(-5,-12)# If that is the question, then the form of the answer will depend on whether you call the second point #(x,f(x))# or #(a,f(a))# or somthing similar of call it #(-5+h, f(-5+h))# For #(x,f(x))#, we find slope: #m = (f(x)-(-12))/(x-(-5)) = (x^2-x-42 + 12)/(x+5) = (x^2-x-30)/x+5)# # = ((x-6)(x+5))/(x+5) = x-6# (for #x != -5#) For #(-5+h, f(-5+h))#, we find slope: #m=(f(-5+h)-(-12))/((-5+h)-(-5))# # = ((-5+h)^2-(-5+h)-42+12)/(-5+h+5)# # = (25-10h+h^2+5-h-42+12)/h# # = (-10h+h^2-h)/h = -11+h# (for #h != 0#) The general equation of a secant line to the curve that includes the point #(7, 0)# is found by similar methods.

Luke Alvarez · Answer

undefined To find the slope of the secant line between two points on a function $ f(x) $, you use the formula:

$$ \text{Slope} = \frac{f(x_2) - f(x_1)}{x_2 - x_1} $$

Given the points $(-5, -12)$ and $(7, 0)$, the coordinates are $ (x_1, f(x_1)) = (-5, -12) $ and $ (x_2, f(x_2)) = (7, 0) $, respectively.

Plug the values into the formula:

$$ \text{Slope} = \frac{0 - (-12)}{7 - (-5)} $$

$$ \text{Slope} = \frac{12}{12} $$

$$ \text{Slope} = 1 $$

So, the slope of the secant line of $ f(x) = x^2 - x - 42 $ at the given points is $ 1 $.