Question

How do you find the slope of a tangent line to the graph of the function `f(x) = x^2-2x` at the point (a,f(a))?

Answer 1

`y=(2a-2)x-a^2`

Explanation:

`f(x)=x^2-2x`

When `x=a => f(a)=a^2-2a`

We get the slope at the tangent at any point by using the derivative,
`f'(x)=2x-2`
When `x=a => f'(a)=2a-2`

So, the tangent passes through the point `(a,a^2-2a)` and has slope `m=2a-2`.

Using `y-y_1=m(x-x_1)` the tangent equation is given by:

`y-(a^2-2a)=(2a-2)(x-a)`
`:. y-a^2+2a=2ax-2a^2-2x+2a`
`:. y=2ax-a^2-2x`
`:. y=(2a-2)x-a^2`