(05.01)Sarah has been running a dog-walking business since 2010. She walks dogs twice a day, takes them to the park, and returns them to their homes. Each year, she has increased her fee by the same amount. The table shows what Sarah charged each customer for two given years of her business: Year Annual Dog-walking Fee 2010 $350 2014 $750 A. What is the rate of change and initial value for Sarah’s business? How do you know? B. Write an equation in slope-intercept form to represent the fees that Sarah charges each year.
I think the rate of change is $200 but I am not sure about the intercept because if the rate of change is 200 the intercept would be negative. IDK I'm just confused. @Vocaloid @563blackghost
Since her dog walking business was started in 2010 the value of $350 is the initial value. This can be represented as `(0,350)`. The table also shows for the year of 2014, since this is exactly 4 years after we can represent this by `(4,750)` so we have our points. \(\bf{A.}\) The rate of change is found by finding the slope between two points. So we plug this in our slope formula. \(\large\bf{slope=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}}\) Plug in your points to find the slope. What is the slope?
100? I was thinking two years for some reason
Yes it would be 100.
(:
The intercept is the initial value or the value of when the business began.
\(\bf{B}\) So you would plug this into slope-intercept form. \(\large\bf{y=mx+b}\)
y = 100x + b i am confused on the B part
`b` is known as the `y-intercept`. The `y-intercept` is where the line of an equation passes through the y-axis so the point is always `(0,#)`. Since the business began in 2010 and the table includes the value of the year 2010 then the value is our `y-intercept`.
OMG! i should have known that! so. y= 100x +2010?
2010 is the year, it would be value of the business in 2010.
So it would be $350.
Oh okay! y= 100x + 350
YUP!
thank you!
np ;)
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