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 \[ y+3=-2(x-1) \\ y+3=-2x+2 \\ y=-2x-1 \]

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 \[ y+3=-2(x-1) \\ y+3=-2x+2 \\ y=-2x-1 \]