How do you use the limit definition to find the slope of the tangent line to the graph #y=x^2+2x# at (-3,3)?

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Answer 1 · 2017-01-20T09:34:23

#lim_(x_->-3# #(x-1)=-4#

Let #x_1=-3#

#lim_(x_->-3# #(f(x)-f(x_1))/(x-x_1#

#lim_(x_->-3# #((x^2+2x)-(x_1^2+2x_1))/(x-x_1#

#lim_(x_->-3# #((x^2+2x)-((-3)^2+2(-3)))/(x-(-3)#

#lim_(x_->-3# #((x^2+2x)-(9-6))/(x+3)#

#lim_(x_->-3# #(x^2+2x-3)/(x+3)#

#lim_(x_->-3# #((cancel(x+3))(x-1))/(cancel((x+3))#

#lim_(x_->-3# #(x-1)=-3-1 =-4#