### Quiz 2.1:

Question:

Find the slope of $$f(x)=x^2-4x$$ at $$(1,-3)$$ and write the equation of the tangent line to $$f(x)$$ at that point.

Solution:

$$\frac{\Delta y}{\Delta x}=\frac{((1+h)^2-4(1+h))-(1^2-4(1))}{h} =\frac{1+2h+h^2-4-4h-1+4}{h}=\frac{-2h+h^2}{h}=-2+h,h \neq 0$$
As $$h \rightarrow 0 , \frac{\Delta y}{\Delta x} \rightarrow -2$$
So the slope of the tangent line is $$m=-2$$
And the tangent line touches the curve at the point $$(1,-3)$$
So the equation of the tangent line is $y+3=-2(x-1) \\ y+3=-2x+2 \\ y=-2x-1$