# 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#