Finding β0 and β1 . The equation for a straight line (deterministic) is y = β0 + β1x. If the line passes through the point (0,1), then x = 0, y = 1 must satisfy the equation. That is, 1 = β0 + β1(0). Similarly, if the line passes through the point (2,3), then x = 2, y = 3 must satisfy the equation: 3 = β0 + β1(2). Use these two equations to solve for β0 and β1, and find the equation of the line that passes through the points (0,1) and (2,3).

Step-by-step explanation:

We are given the following in the question:

The line passes through (0,1).

The line passes through (2,3)

Thus, the equation of the line that passes through the points (0,1) and (2,3) is given by:

Slope:

Intercept:

Equation:

is the required equation.

d

step-by-step explanation:

given

y + 3 = - 4(x - 2) ← distribute the parenthesis

y + 3 = - 4x + 8 ( subtract 3 from both sides )

y = - 4x + 5 → d

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.