What is the distance between the vertices of the graphs corresponding to y = (x - 3)2 and y = x2 - 4? a.0 b.1 c.5 d.13

C.5

Step-by-step explanation:

y = (x - 3)^2

This is in the form y= a(x-h)^2 +k  where (h,k) is the vertex

The vertex is at (3,0)

y = x^2 - 4

This is in the form y= a(x-h)^2 +k  where (h,k) is the vertex

The vertex is at (0,-4)

The distance between the points is found by

d = sqrt( (x2-x1)^2 + (y2-y2)^2)

  = sqrt(( 0-3)^2 + (-4-0)^2)

  = sqrt( 9 + 16)

  = sqrt(25)

   = 5

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below:

answer: get jesus in your life

step-by-step explanation: