You are given the dollar value of a product in 2012 and the rate at which the value of the product is expected to change during the next 5 years. Write a linear equation that gives the dollar value V of the product in terms of the year t. (Let t = 0 represent 2010.)
2012 value $950
Rate $350 increase per year
V(t)=??
you know that y=mt+b is the line which starts at b and grows by m for each successive value of t.
So, where does V start (when t=0), and how much does it grow when t increases by 1? That is, for each year?
V starts at 950?
Upon rereading the problem, I must revise my answer. You are given a point in 2012 (where t=2): (2,950)
and also the slope, or change each year. That is, m=350
So, using the point-slope form,
V(t) = 350(t-2)+950
So would that be my final answer or would I have to distribute?
depends how specific your teacher is.
To me, any correct form of the equation will do.
If you teacher specified the slope-intercept form, then yeah, expand it out.
But the point-slope form shows just how you got the equation, not just how well you can manipulate expressions.
Okay thank you. Here is another example with different numbers. Can you tell me if this is correct
You are given the dollar value of a product in 2012 and the rate at which the value of the product is expected to change during the next 5 years. Write a linear equation that gives the dollar value V of the product in terms of the year t. (Let t = 0 represent 2010.)
250t+500
Is this the point slope form as well
2012 value is $550 and rate is $150 increase
150t+250*** my answer
looks good
To write a linear equation that gives the dollar value V of the product in terms of the year t, we can start by finding the initial value in 2010 (t=0).
Given that the value in 2012 is $950, we need to determine the value of the product in 2010. To find this, we can subtract the rate of increase per year from the value in 2012 for each year from 2010 to 2012.
In 2010 (t=0), the value would be $950 - ($350 x 2) since the increase per year is $350 and we are going back by 2 years (from 2012 to 2010).
So the value in 2010 would be $950 - ($350 x 2) = $950 - $700 = $250.
Now that we have the initial value in 2010, we can write the linear equation. The rate of increase per year is $350.
The equation for a linear relationship of the form V(t) = mt + b, where m is the slope (rate of increase) and b is the y-intercept (initial value), can be used.
In this case, m = $350 (rate of increase per year) and b = $250 (initial value in 2010).
Therefore, the linear equation that gives the dollar value V of the product in terms of the year t is:
V(t) = $350t + $250
Where t represents the number of years since 2010.