Question

Find the slope of the function's graph of the given point, then find the equation for the line tangent to the graph there? f(x)=x^2+1 (2,5)

Answer 1

Equation of the tangent is `4x-y-3=0`

Explanation:

Observe that as `f(x)=x^2+1`, `5=2^2+1` and hence `(2,5)` lies on the curve `f(x)=x^2+1`

As `f(x)=x^2+1`, `f'(x)=(df)/(dx)=2x`

and therefore slope of the function's graph of the given point `(2,5)` is `2xx2=4`

hence equation of the tangent having slope `4` and passing through `(2,5)` is

`y-5=4(x-2)` or `4x-y-3=0`

graph{(x^2+1-y)(4x-y-3)=0 [-9.38, 10.62, -2.08, 7.92]}