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!
(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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