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#?
1 Answer
We have:
# y=x^2+3x #
When
# y =3^2+3(3) #
# \ \ =9+9 #
# \ \ =18 #
When
# y = (3+h)^2+3(3+h) #
# \ \ = 9+6h+h^2 + 9 + 3h #
# \ \ = h^2+9h+18 #
So the gradient of the secant line is given by:
# 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
Conclusion
Note that as
# m_(tan) = lim_(h rarr 0) m_(sec) #
# " " = lim_(h rarr 0) (h+9) #
# " " = 9 #
It should be clear that this final result comes as a direct result of the definition of the derivative. We can also use our knowledge of Calculus to validate this, as:
# y=x^2+3x => dy/dx = 2x + 3#
And so, when
# [dy/dx]_(x=3) = 2(3) + 3 = 9#