Solve each problem. During the course of ayear, the number of volunteers available to run a food bank each month is modeled by V(x), where V(x)=2x^2-32x+150 between the months of January and August. Here x is time in months, with x=1 representing January. From August to December, V(x) is mod-eled by V(x)=31x-226.

Find the number of volunteers in each of the following months.

Sketch a graph of y=V(x) for January through December. In what month are the fewest volunteers available?

Question

Solve each problem. During the course of ayear, the number of volunteers available to run a food bank each month is modeled by V(x), where V(x)=2x^2-32x+150 between the months of January and August. Here x is time in months, with x=1 representing January. From August to December, V(x) is mod-eled by V(x)=31x-226.

Find the number of volunteers in each of the following months.

Sketch a graph of y=V(x) for January through December. In what month are the fewest volunteers available?

Accepted Answer

Welcome back. I am so glad you're here. We are told in a certain university, the number of students who participated in the yearly fundraising activity from 2011 to 2018 can be modeled by them. Were given the equation, the function A of X equals three X squared minus 52 X plus 255 where X equals one represents the year from 2018 to 2022. The number of students is represented by A FX equals 42 X minus 305. We are asked to graph Y equals a FX for the years 2011 to determine the year in which the number of students who participated is the least. All right. So what do we know here? Well, we're told that X equals one is the year 2011. So our X axis, our values are yours. We know that our domain, the X values that are possible is from 2011 to 22. Those are the year's domain and X values. What does that translate to? We can create a chart? Where starting in the year 2011, that's an X value of one. And we don't yet know the corresponding Y value, but we can go all the way from 2011, 2012. The X value would be 2, 2013. X is 3 2014. X is four, 2015, X is 5 2016. X is 6, 2017. X is seven, 2018, X is eight. But also in the year 2018, we're using this second function. So again, 2018 is eight again, but now 2019 is 9, 2020 is 10, 2021 is 11. And our final X value 2022 will be 12. So our domain for our X values is from 1 to 12. How about our Y axis? Well, our Y values or our A FX, that's the number of students who participated in the fundraising. So our Y values are students. What's our range for students? Well, those are only going to be positive numbers. We're going to have positive students. We're not going to have negative students. So our range is going to be from zero to positive infinity for our Y values. And I suppose I can write the X values rain or domain in interval notation as well. So that's open bracket, comma or one comma 12 close bracket. OK. So to graph this, because everything here is positive, we can draw just the first quadrant for our graph. So we'll have a vertical Y axis, a horizontal X axis. And it's only showing the first quadrant. We know that our domain for our X values is from 1 to 12. So we can draw these 123456789, 10, 11 and then 12. And for our Y values, our range going from zero. Well, that goes up to infinity. So we don't know what our um the height of the realistic values that we're talking about here is. So how can we figure out what our range is going to be? Well, what we can do is we can plug in our, all of our X values and determine all of the corresponding Y values. And that would let us know our range. It's just going to be from zero to whatever the top of that, the highest number in our Y values is. That's one way to figure it out. We have other clues looking at our functions. We can analyze these functions a little bit from 2011 to 2018. That function is a Parabola that's got this X squared. Now, it's not in our standard Parabola format, but we can see that our, a value the one in front of the X squared, our A is greater than zero. So this is a Parabola that's going to open upward. And so it'll look something like this when we drive it. And we also know that our, why intercept, even though our Y intercept is outside of our domain, it can give us a clue as to what's going on our Y intercept. We would find that by finding, well, this is an F, this is a, a of zero instead of A, of X plugging zero in for X three, X squared, that's three multiplied by zero squared. That's nothing minus 52 multiplied by zero. That's nothing. So our Y intercept would be at 255 if that was within our domain. So if we've got a parabola that intercepts with the Y access around 255 and then heads downward. That gives us a clue for our range that possibly we're talking about 255 being toward the max for our range. So that gives us some clues. We know that the second function looking at that is a line that's in the format of Y equals M X plus B B is our Y intercept. B would be a negative 305. Well, again, that's outside of our range or outside of our domain. The domain for this function is only from 8 to 12. And so plugging in for zero, there gives us a Y intercept well, outside of our range. But it could be looking something kind of like this. It will go down and intercept with the Y axis and the negative values and the slope M equals a positive 42. So that gives us some clues. But the most straightforward thing here would be to just plug in our X values into the functions, figure out the corresponding Y values and then plot all of those points. This is something that I've done ahead of time. So I'm just going to fill them out. You're welcome to pause the video and figure out what those corresponding Y values are. When we plug one into our first function, our Y value is 206. If we're plotting that from the origin, we go to the right one unit and then up 206 units. So yeah, when we said 255 was the Y intercept. This is bit below it. We can label right here. We've got 202 106 would be a little bit above that. So let's have our Y axis labels go up by fifties. So from the origin, we can go up 50 100. I'm not spacing that too well, 50 can be higher than that. Oh, no, I lost my lines. There we go. 50 150 200. All right, let's find out what our other Y values are. When X is two Y is from the origin to the right two units and then up 163 units when X is three Y is 126 from the origin to the right three units up 126 units when X is four, Y is 95 from the origin to the rate four units up 95 in 2015. When X is five, the Y value is 70. So we had 70 students that year from the origin to the rate six units and or excuse me to the right five units and up units in 2016, X is six, Y is 51 from the origin to the right six units. Up 51 units. 2017, X seven Y is from the origin to the right seven units and up 38 units. And in 2018, X is eight and Y is 31 from the origin to the right eight units up 31 units. And then we can see our parabola. So if we connect these points, this curves down and it looks like it's approaching its vertex. OK? Now we can switch to the other function for the years from 2018 to 2022. We plug eight in for X into that second function. We get 31 again. So these are connected, we can put our point again at 31. Now we said this is straight line, well, straight line, we only need two points so we can go down to put 12 in for X and we get a Y value of 199. So still within our range that we set up for the Y values going to the right 12 units and then going up units. And then we connect 8 31 with 12, 199 straight line for that one. And there we have our graph. Now, the only other question we're asked is, what's the year in which this number of students participated is the least? And the lowest point on our graph is in this eighth year here, which is 2018 when only 31 students participated. All right, there's our graph and there's our answer. Well done. We'll catch you on the next one.

Verified Solution

Key Concepts

Quadratic Functions

Piecewise Functions

Graphing Functions