WILL UPVOTE AND WOULD LIKE AN EXPLANATION AS WELL PLEASE!!
The table below represents the velocity of a car as a function of time:

Time
(hour)
x Velocity
(miles/hours)
y
0 50
1 52
2 54
3 56
Left side is X and right side is Y

Part A: What is the y-intercept of the function, and what does this tell you about the car? (4 points)

Part B: Calculate the average rate of change of the function represented by the table between x = 1 to x = 3 hours, and tell what the average rate represents. (4 points)

Part C: What would be the domain of the function if the velocity of the car was measured until it reached 60 miles/hour and the car does not change motion? (2 points)

Respuesta :

(A): When it asks for a y-intercept, it is basically asking what is the value of y when x=0. In this case, when x=0, y=50. So the y-intercept is 50.

(B): The avg. rate of change is the same as asking for slope. So we want to use slope formula -> [tex]m= \frac{y2-y1}{x2-x1} [/tex]. Using the values in your table -> [tex]m = \frac{56-52}{3-1} = \frac{4}{2} =2 [/tex]

(C) You can generate a linear function with a y-intercept and a slope using the following formula -> y= (average rate of change)*x + y-intercept ->[tex]y= 2x+50[/tex] Then set y=60 and solve for x. Your domain is [0, solved value] with square brackets since you are including the endpoints.

Hope that helped!

Answer:

(A): When it asks for a y-intercept, it is basically asking what is the value of y when x=0. In this case, when x=0, y=50. So the y-intercept is 50.

(B): The avg. rate of change is the same as asking for slope. So we want to use slope formula -> . Using the values in your table -> 

(C) You can generate a linear function with a y-intercept and a slope using the following formula -> y= (average rate of change)*x + y-intercept -> Then set y=60 and solve for x. Your domain is [0, solved value] with square brackets since you are including the endpoints.

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