# The derivative of f(x)=x^2-4x+5 is f'(x)=2x-4. Find the equation of the tangent line to the graph...

## Question:

The derivative of {eq}f(x)=x^2-4x+5 {/eq} is {eq}f'(x)=2x-4. {/eq} Find the equation of the tangent line to the graph of {eq}f(x) {/eq} at the point {eq}(1,2). {/eq}

## The Tangent Line to a Curve:

The concept of tangent line and slope is used to solve many problems in mathematics. We are given equation of the curve involves a quadratic function and we need to find out the equation of the tangent line of the given curve.

The slope of the tangent line is the first derivative of the given function and the equation of the tangent line in point-slope form is {eq}y=mx+b {/eq} .

We are given the curve {eq}f(x) = x^2-4x+5 {/eq}

The derivative is given, i.e {eq}f'(x) =2 x-4. {/eq} . Hence the slope of the given curve is {eq}m= 2x-4 {/eq}

Plug in {eq}x=1 {/eq} then {eq}m= 2\cdot 1-4=2-4=-2 {/eq}

Now the line equation {eq}y = mx+b {/eq}

Plug in {eq}m=- 2 {/eq} and the point {eq}\left ( 1\ , \ 2 \right) {/eq} we'll get:

{eq}2=-2 \cdot 1 +b {/eq}

{eq}\Rightarrow b-2= 2 {/eq}

{eq}\Rightarrow b= 2+2 {/eq}

{eq}\Rightarrow b= 4 {/eq}

Therefore, the equation of the tangent line to the curve {eq}y = x^2-4x+5 {/eq} when {eq}(1,2) {/eq} is {eq}{\boxed{ y =- 2x+4 .}} {/eq}