Find an equation of the plane that passes through the point (-1,1/2,3) with normal vector [1,4,1]

x + 4y + z = 4

Step-by-step explanation:

The standard equation is of the form, Ax + By + Cz = 0

The equation of the plane that passes through the point(-1, 1/2, 3)  and is perpendicular to [1, 4, 1] is

A = 1, B = 4, C = 1

x = -1; y = 1/2; z = 3

A(x-(-1)) + B(y-1/2) + C(z-3) = 0

(x+1) + 4(y-1/2) + (z-3) = 0

(x+1) + (4y-2) + (z-3) = 0

x + 1 + 4y -2 + z -3 = 0

x + 4y + z - 4 = 0

x + 4y + z = 4

answer: rolling a number cube and flipping a coin.

step-by-step explanation:

step-by-step explanation:

we have

using a graphing tool

see the attached figure

this is a quadratic equation (vertical parabola) open up

the vertex is a > vertex

the domain of the function is all real > interval (-∞,∞)

the range of the function is the > [-6.125,∞)

the function has two x->   and

the function is increasing over the > (0.25,∞)

the function is decreasing over the > (-∞,0.25)