Find the equation of two tangent lines to the graph of f(x)=x^2 + 1 that pass through the (0,0) point.?

Question

Please show all steps.

Answer 1

f(x) = x^2 + 1
f '(x) = 2x
At x = 1 :
f '(1) = 2(1) = 2
The line y = 2x passes through the point (1, 2) and is tangent to f(x) at that point.
Also by symmetry, y = -2x passes through the point (-1, 2) and is tangent to f(x) at the point (-1, 2)

Answer 2

Open question instruct me yet another » Open question instruct me yet another » Ind equations of the tangent lines to the graph f(x)=x/(x-a million) that pass in the time of the factor (-a million,5).? y=x/(x-a million) dy/dx=[(x-a million)(d/dx)(x)-x(d/dx)(x-a million)]÷(x... dy/dx=[(x-a million)(a million)-x(a million)]÷(x-a million)² dy/dx=(x-a million-x)÷(x-a million)² dy/dx=-a million/(x-a million)² [dy/dx]at(-a million,5)=[-a million/(-a million-a million)² dy/dx]at(-a million,5)=[-a million/(-2)² dy/dx]at(-a million,5)=(a million/4) Equation of tangent; [y-y?]=(dy/dx)[x-x?] y-5=(/4)[x+a million] 4y-20=x+a million x-4y+a million+20=0 x-4y+21=0