Find a vector equation for the line through the point P = (4, 2, 4) and parallel to the vector v = < -2, -5, -3 > . Assume r(0) = 4i + 2j + 4k and that v is the direction vector of the line. R(t) = Rewrite this in terms of the corresponding parametric equations for the line. x = y = z =

The parametric equations are

 x = - 2 t + 4 , y = - 5 t + 2 , z = - 3 t + 4

Step-by-step explanation:

Step(i):-

Given point P = ( 4,2,4)

Given vector is   -2 i - 5 j - 3 k

The vector equation of a line through the point a⁻ and parallel to b⁻ is

r⁻ = a⁻ + t b⁻  

Given r⁻ = 4 i + 2 j + 4 k

r⁻ = 4 i + 2 j + 4 k + t (-2 i - 5 j - 3 k)

Step(ii):-

The Cartesian form

⇒ x - 4 = -2 t

⇒ x = - 2 t + 4

⇒ y - 2 = - 5 t

⇒ y = - 5 t + 2

z - 4 = -3 t

z = - 3 t + 4

Conclusion:-

The parametric equations are

 x = - 2 t + 4 , y = - 5 t + 2 , z = - 3 t + 4

the graph in the attached figure

step-by-step explanation:

we have

using a graphing tool

the graph in the attached figure

the roots are ( x-intercepts) x=-0.5,x=1,x=3

the y-intercept is the point (0,3)

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.