Show that the gradient of the secant line to the curve #y=x^2+3x# at the points on the curve where #x=3# and #x=3+h# is #h+9#?

Question

Show that the gradient of the secant line to the  curve #y=x^2+3x# at the points on the curve where #x=3# and #x=3+h# is #h+9#?

Paisley Johnson · Accepted Answer

We have: # y=x^2+3x # When #x=3# we have: # y =3^2+3(3) #
# \ \  =9+9 #
# \ \  =18 # When #x=3+h# we have: # y = (3+h)^2+3(3+h) #
# \ \  = 9+6h+h^2 + 9 + 3h #
# \ \  = h^2+9h+18 # Hence, the gradient of the secant line can be found using: # m_(sec)  = (Delta y)/(Delta x) # # "    " = (y_(3+h) - y_h)/((3+h)-(3)) # # "    " = ((h^2+9h+18) - (18))/(3+h-3) # # "    " = (h^2+9h)/(h) # # "    " = h+9 \ \ \ # QED In summary Note that as #h rarr 0# then the secant line becomes the tangent, so in the limit we have: # m_(tan) = lim_(h rarr 0) m_(sec) # # "    " = lim_(h rarr 0) (h+9) # # "    " = 9 # It should be evident that the definition of the derivative leads directly to this final result. We can confirm this by applying our calculus knowledge, as follows: # y=x^2+3x => dy/dx = 2x + 3# And so, when #x=3#, we have: # [dy/dx]_(x=3) = 2(3) + 3 = 9#

Isaiah Allen · Answer

undefined To find the gradient of the secant line, we need to calculate the slope between two points on the curve.

Let's start by finding the coordinates of the two points on the curve. When x = 3, the corresponding y-value can be found by substituting x = 3 into the equation y = x^2 + 3x:

y = (3)^2 + 3(3) = 9 + 9 = 18

Now, let's find the coordinates of the second point when x = 3 + h:

y = (3 + h)^2 + 3(3 + h) = 9 + 6h + h^2 + 9 + 3h = h^2 + 9h + 18

Now that we have the coordinates of the two points, we can calculate the slope (gradient) of the secant line using the formula:

Gradient = (change in y) / (change in x)

(change in y) = (h^2 + 9h + 18) - 18 = h^2 + 9h
(change in x) = (3 + h) - 3 = h

Therefore, the gradient of the secant line is:

Gradient = (h^2 + 9h) / h = h + 9