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

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Answer 1 · 2017-01-20T09:34:23

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

Let $x_1=-3$

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

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

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

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

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

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

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

How do you use the limit definition to find the slope of the tangent line to the graph y=x^2+2xy=x2+2xy=x^2+2x at (-3,3)?