Problem

Solve each problem. See Examples 6 and 7. The pe…

Problem 43 Easy Difficulty

Solve each problem. See Examples 6 and 7.
The total receipts from individual income taxes by the U.S. Treasury in the years $2000-2007$ can be modeled by the quadratic function defined by
$$
f(x)=22.88 x^{2}-141.3 x+1044
$$
where $x=0$ represents $2000, x=1$ represents $2001,$ and so on, and $f(x)$ is in billions of dollars. (Source: World Almanac and Book of Facts.)
A. since the coefficient of $x^{2}$ given in the model is positive, the graph of this quadratic function is a parabola that opens up. Will the $y$ -value of the vertex of this graph be a maximum or minimum?
B. In what year during this period were total receipts from individual taxes a minimum? (Round down for the year.) Use the actual $x$ -value of the vertex, to the nearest tenth,
to find this amount.

Answer

(a) minimum
(b) $2003 ; \$ 825.8$ billion

Related Courses

Precalculus

Algebra

Beginning and Intermediate Algebra

Chapter 11

Quadratic Equations, Inequalities, and Functions

Section 7

More about Parabolas and Their Applications

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Video Transcript

Let's take a look at our function. F of X equals 22.88 X squared minus 141.3 x plus 1044. Now, by looking at my X square term, I can see that this is a parabola. It has a positive coefficient. So this parable a is upward facing, which means that my Vertex is going to give me the minimum value of this function. So what is the Vertex? What value of X gives me the minimum amount for the individual taxes? Well, to find that I'm going to need to complete the square every term with an ex, I'm going to pull out and put together every term without an X. In this case, 1044. I'm gonna push over to the side, Out of our way. Just a note. Our final answer. We will need to be rounding to the nearest tents, but I don't want around too soon. That could introduce a lot of rounding errors in the problem. So I'm going to carry 4 to 5 decimal places throughout this problem. Hopefully that will give me the least rounding error possible. So first X squared needs to have a coefficient of one. So I'm going to factor out a 22.88 from every term with an accident. So that's going to give me X squared minus 6.1757 x again. I'm going to about 4 to 5. Decimal places would leave a nice big gap in those parentheses and I'll have plus 1044 over on the side. Okay, To complete the square, we look at the X term. I'm going to take half of that coefficient negative. 3.8785 square that I'm gonna plug that back into my equation. That gives me a plus 9.5348 Now I can't just add a number toe a function, so I'm going to subtract 9.5348 as well, by adding it's attracting the same number. I haven't changed the overall value of this function. All the pieces I need to complete the square are X squared minus 6.1757 x plus 9.5348 I need to move this minus 9.5348 outside of the parentheses. But don't forget, everything in the parentheses is being multiplied by 22.88 So when I move it outside of the parentheses, what I'm really moving out is minus 218 0.156 to which I will combine with the 1044 that's already out there. When I do this, completing the square gives me 22.88 times X minus three point. Oh, hope sorry. 08785 squared No, plus 825.8438 So my Vertex can be found at the point 3.8785 8.825 point 8438 Now let's evaluate this Vertex. In light of the problem, the X coordinate tells me the year past the year 2000 so and I'm supposed to round the year down surrounding that down gives me an X value of three. Which means that my year is 2003. And what amount do I have there? Well, im supposed, that's my y coordinate. Supposed to round at the nearest tense, so I get 825.8 $1,000,000,000

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