medal-----I just need help figuring out this question not just the answer please
The table below represents the velocity of a car as a function of time:
Time Velocity
(hour) (miles/hours)
x y
0 50
1 52
2 54
3 56
Part A: What is the y-intercept of the function, and what does this tell you about the car?
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.
Part C: What would be the domain of the function if the velocity of the car was measured until it reached 60

- anonymous

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- anonymous

Help please

- anonymous

Could you please help me if you can @Nnesha or @whpalmer4 or if you know someone you can tag to help

- whpalmer4

What is the definition of the \(y\)-intercept?

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## More answers

- anonymous

yes here

- whpalmer4

So, what is the definition of the \(y\)-intercept?

- anonymous

it is the point on a graph when the x is equalled to 0....?

- whpalmer4

Well, it's the value of \(y\) at the point on the graph where \(x=0\). Also the point where the graph crosses the \(y\)-axis.

- whpalmer4

So if we graphed the data in the table, would the \(x\)-axis be time, or velocity?

- anonymous

x wold be time because the velocity would be y

- whpalmer4

good. so now you can tell me the answer to part A...

- anonymous

so when x is 0 y is 50

- whpalmer4

Yes. And as \(x\) represents time, that's the initial velocity of the car, right? 50 miles/hour

- anonymous

yes so all i have to write would be.... the y intercept would be 50 if x is 0 im not sure how to word it

- whpalmer4

How about "the y-intercept of the function is 50 miles/hour"?

- anonymous

ok ..... :) but it also says what does this tell you about the car? i dont know how to figure that out or does that answer the entire question ? @whpalmer4

- whpalmer4

If you look back a few responses, I think you will find that I made a comment about what that represents...

- whpalmer4

The data in the table shows the car starting at 50 mph and going faster with each passing hour, right?

- anonymous

yes by 2 miles per hour

- whpalmer4

no, but close: it changes by 2 miles per hour per hour...

- whpalmer4

or miles/hour^2

- anonymous

yes :) your right

- whpalmer4

no, "you're right" :-)

- whpalmer4

On to part B! How do you find the average rate of change?

- anonymous

average rate of change for an function f(x) is intervals from a to b so the formula is f(b)-f(a) over b-a

- whpalmer4

okay, do you understand how to translate your table values into a,b,f()?

- anonymous

@whpalmer4 i was reading my notes from class would F(a) be x and f (b) be y

- anonymous

so would it be from the table 52-50 over 1-0 ??? not sure

- whpalmer4

Well, \(x\) takes on the values of \(a\) and \(b\) and \(y = f(x)\)
So at \(x=1 \text{ hour}\) we have \(a = x = 1\text{ hour}\) and \(f(1) = \) what?

- anonymous

52

- whpalmer4

Yes. so \(a = 1\) and \(f(a) = 52\)
Now how about \(b\) and \(f(b)\)?

- anonymous

b =2 and f(b)=2

- anonymous

f(b) =54 oops

- whpalmer4

why does b = 2? and f(b) is the value of the function at x=b=2, which doesn't appear to be 2 when I read the table...
Here's the initial problem again:
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.
^^^^^^^^^

- whpalmer4

well, my attempt to underline "\(x=3\)" didn't work so well!

- anonymous

ohhh So b =3 and f(b)=56

- whpalmer4

yes! we are trying to find the average rate of change over that 2 hour span, so \(a\) is the value of \(x\) or \(t\) at the start, and \(b\) is the value of \(x\) or \(t\) at the end, and \(f(a)\) and \(f(b)\) are the values from the table.

- whpalmer4

So our average rate of change between \(a\) and \(b\) is as your formula states:
\[\text{avg rate of change} = \frac{f(b)-f(a)}{b-a}\]

- whpalmer4

Can you plug in the values?

- anonymous

56-52/ 3-1=4/2 =2

- whpalmer4

Okay, if you are going to write equations on one line like that, you MUST use parentheses appropriately to show what isn't being shown by position!
what you wrote is actually equal to this:
\[56 - \frac{52}{3} - 1 = \frac{4}{2}\]
I know that isn't what you meant.
(56-52)/(3-1) = 4/2 would unambiguously convey the intended meaning

- whpalmer4

What will the units for your answer be? miles? miles per hour? something else?

- anonymous

oh ok thanks for that tip and my units would be miles per hour or miles/hour

- anonymous

so ithe change of rate would be 2 miles per hour

- whpalmer4

right number, wrong unit
\[\frac{56 \text{ miles/hr} - 52 \text{ miles/hr}}{3\text{ hr} - 1\text{ hr}} = \frac{4\text{ miles/hr}}{2\text{ hr}} = 2 \text{miles/hr}^2\] \]

- anonymous

can you explain why is it hr^2

- anonymous

I dont know how to get the little 2 sorry

- whpalmer4

right? if we continue accelerating at this rate, for every hour that passes, we will increase our speed by 2 miles/hour. If we accelerate for 10 hours, we will add \[10\text{ hours}*2\frac{\text{ miles}}{\text{hour*hour}} = 10\cancel{\text{ hours}}*2\frac{\text{ miles}}{\text{hour}*\cancel{\text{hour}}} \]
\[= 20 \frac{\text{ miles}}{\text{hour}}\] to our speed

- anonymous

ohhh yes ........ :0

- anonymous

thanks for explaining it

- anonymous

my mom says we have to go to church can i see if your on later to finish :)

- whpalmer4

the way it "builds up" in physics terminology:
\[x = v t\]distance = speed * time
\[v = a t\]velocity = acceleration * time

- whpalmer4

Sure, say hi to the big guy for me :-)

- anonymous

lol i will thank you so much

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