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))?

Question

Scarlett Allison · Accepted Answer

Slope #= h+17#

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 #

Christopher Allers · Answer

undefined To find the slope of the secant line between the points $(6, f(6))$ and $(6 + h, f(6 + h))$ on the function $f(x) = x^2 + 5x$, we first need to find the corresponding function values at these points.

$f(6) = (6)^2 + 5(6) = 36 + 30 = 66$

$f(6 + h) = (6 + h)^2 + 5(6 + h) = 36 + 12h + h^2 + 30 + 5h = h^2 + 17h + 36$

Now, we can use the formula for the slope of a secant line:

$$ \text{Slope} = \frac{\text{Change in } y}{\text{Change in } x} = \frac{f(6 + h) - f(6)}{(6 + h) - 6} $$

$$ = \frac{(h^2 + 17h + 36) - 66}{h} $$

$$ = \frac{h^2 + 17h - 30}{h} $$

This expression represents the slope of the secant line between the two given points on the function $f(x) = x^2 + 5x$.