Respuesta :

irspow
d^2=(dx)^2+(dy)^2  but since we are finding the distance from the origin we just have:

d^2=x^2+y^2  and since y=4x+5:

d^2=x^2+16x^2+40x+25

d^2=17x^2+40x+25

dd^2/dx=34x+40

d2d^2/d2x=34

since the acceleration is always a constant positive, when the velocity is equal to zero, d^2 is at an absolute minimum...

dd^2/dx=0 when 34x+40=0, 34x=-40, x=-40/34, x=-20/17

y(-20/17)=-80/17+5=(-80+85)/17=5/17

So the point on the line y=4x+5 that is closest to the origin is:

(-20/17, 5/17)  or if you like approximations:

(-1.18, 0.29)