Question

The points (0, 1) and (1, 4) are contained in the graph of an equation with only two variables, I and y. Select all of the true statements.

a. There is exactly one equation in the form y = mx + b that contains these points.
b. There is exactly one equation in the form y = a. bˣ that contains these points.
c. There is more than one equation in the form y=a.bˣ that contains these points.
d. There are no equations in the form y = a.bˣ that contain these points.
e. There are two equations in the form y = mx + b that contain these points.

Answer 1

The two points (0, 1) and (1, 4) lie on the graph of exactly one linear equation y = 3x + 1 and one exponential equation y = 4^x.

Step-by-step explanation:

When examining the points (0, 1) and (1, 4), we can see that they can be part of a graph of a linear equation in the form y = mx + b. To find the equation of the line that contains these points, we first calculate the slope (m) using the two points. The slope (m) is determined by the change in y divided by the change in x, which is (4-1)/(1-0) = 3. Therefore, the slope m = 3.

Next, we use one of the points to solve for the y-intercept (b). Using the point (0, 1) where x=0, y=1, we can plug these values into the equation y = mx + b to get 1 = 3*0 + b. This simplifies to 1 = b. Thus, our equation is y = 3x + 1.

Examining the quadratic form y = such an equation exists, it should satisfy both points given. Using the points (0, 1) and (1, 4), we set up two equations: 1 = a ⋅ b0 (since anything raised to the power of 0 is 1, this simplifies to 1 = a) and 4 = a ⋅ b1 or 4 = b when we replace 'a' with 1. This means there is one exponential equation y = 1 ⋅ 4x or y = 4x which passes through both points.

Hence, the correct answers are: a. There is exactly one equation in the form y = mx + b that contains these points and b. There is exactly one equation in the form y = a ⋅ bx that contains these points.