If #y=x^2+3x# then find the slope of the secant line at #x=3# and #x=3+h#. What happens as #Deltax rarr 0#? Take #lim_(Deltax-0) (f(x+Deltax) - f(x))/(Deltax)#.

Question

If #y=x^2+3x# then find the slope of the secant line at #x=3# and #x=3+h#.  What happens as #Deltax rarr 0#? Take #lim_(Deltax->0) (f(x+Deltax) - f(x))/(Deltax)#.

Emma Aldridge · Accepted Answer

Well, I assume you know how to find a slope in general, just not how to apply it here:

#"Slope" = (Deltaf(x))/(Deltax) = (f(x_f) - f(x_i))/(x_f - x_i)#

If you want to estimate the slope of #f(x) = x^2 + 3x# at #x = 3#, you can pick points #(x,f(x))# around/near #x = 3# so that you can find the slope of a line that touches #f(x)# at #x = 3#. The closer we are to #x = 3# with our chosen points, the more accurate we will be.

Say we chose #x_f = 3.1# and #x_i = 2.9#. This gives:

#f(3.1) = (3.1)^2 + 3(3.1) = 18.91 = f(x_f)#

#f(2.9) = (2.9)^2 + 3(2.9) = 17.11 = f(x_i)#

So, the slope around #x = 3# with a spread of #Deltax = 3.1 - 2.9 = 0.2# gives:

#color(blue)("Slope") ~~ (18.91 - 17.11)/(3.1 - 2.9)#

#~~ color(blue)(9)#

graph{(y - x^2 - 3x)(y - 18 - 9(x - 3)) = 0 [-10, 10, -9.65, 48.9]}

The above graph shows the straight line that represents the slope at #x = 3#.

The true slope at #x = 3# is found by taking the derivative, which is another way of saying "let #Deltax -> 0# and let's see what the slope becomes".

#lim_(Deltax->0) (f(x_f) - f(x_i))/(x_f - x_i)#

#= (f(x_i + Deltax) - f(x_i))/(x_f - x_i)#

A simple way to take a derivative, #(df)/(dx)#, of a polynomial is with the power rule:

#d/(dx)[x^n] = nx^(n-1)#

So, for #f(x) = x^2 + 3x#:

#(df)/(dx) = 2 cdot x^(2-1) + 3(1 cdot x^(1-1))#

#= 2x + 3#

And at #x = 3#, the derivative gives the slope at #x = 3#:

#color(blue)(|[(df)/(dx)]|_(x=3)) = 2(3) + 3 = color(blue)(9)#

So, it turns out that we chose our #Deltax# in such a way that our estimated, average slope was the actual slope.

Emma Aldridge · Answer

Slope = #9#

I will answer this question with reference to a similar relevant question that you asked:

In this similar question:

https://tutor.hix.ai

we were asked to show that the expression for the slope of the secant line through #y = x^2+3x# at #x =3# and #x = 3+h# is:

# m_(sec) = h+9 #

So if we take small values of #h# (eg #h=0.01#) then we get a reasonable estimate of the slope of the tangent at #x=3#, in this case, our estimate would be slope#=0.01+9=9.01#.

In this question, I further demonstrated that as #h rarr 0# then the secant line approaches the tangent with more and more accuracy, and, so in the limit we have:

# m_(tan) = lim_(h rarr 0) m_(sec) #

# " " = lim_(h rarr 0) (h+9) #

# " " = 9 #

Luke Alvarez · Answer

undefined To find the slope of the secant line at x=3 and x=3+h, we need to calculate the difference in y-coordinates divided by the difference in x-coordinates.

First, let's find the y-coordinates:
For x=3, y = (3)^2 + 3(3) = 9 + 9 = 18.
For x=3+h, y = (3+h)^2 + 3(3+h) = 9 + 6h + h^2 + 9 + 3h = h^2 + 9h + 18.

Now, let's find the difference in y-coordinates:
f(x+Deltax) - f(x) = (h^2 + 9h + 18) - 18 = h^2 + 9h.

Finally, let's find the difference in x-coordinates:
Deltax = (3+h) - 3 = h.

Now, we can calculate the slope of the secant line:
lim_(Deltax->0) (f(x+Deltax) - f(x))/(Deltax) = lim_(h->0) (h^2 + 9h)/h = lim_(h->0) (h + 9) = 9.

As Deltax approaches 0, the slope of the secant line approaches 9.

If #y=x^2+3x# then find the slope of the secant line at #x=3# and #x=3+h#. What happens as #Deltax rarr 0#? Take #lim_(Deltax-&gt;0) (f(x+Deltax) - f(x))/(Deltax)#.

If #y=x^2+3x# then find the slope of the secant line at #x=3# and #x=3+h#. What happens as #Deltax rarr 0#? Take #lim_(Deltax->0) (f(x+Deltax) - f(x))/(Deltax)#.