June 2023 Algebra 2 Regents, Part III

This exam was adminstered in June 2023.

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Algebra 2 June 2023

Part III: Each correct answer will receive 4 credits. Partial credit can be earned. One computational mistake will lose 1 point. A conceptual error will generally lose 2 points (unless the rubric states otherwise). It is sometimes possible to get 1 point for a correct answer with no correct work shown.


33. Patricia creates a cubic polynomial function, p(x), with a leading coefficient of 1. The zeros of the function are 2, 3, and -6. Write an equation for p(x).

Sketch y = p(x) on the set of axes below.

Answer:

You are give the zeroes of the function and you need to have a 1 leading coefficient. You can write the function in factored form. You do not need to multiply it.

p(x) = (x - 2)(x - 3)(x + 6)

If you aren't sure what this looks like, you can put it in your graphing calculator and look at it and the table of values.

When a cubic function has a positive leading coefficient, it starts at negative infinity and climbs to positive infinity. The end behavior is that as x goes to negative infinity, y goes to negative infinity, and when x goes to positive infinity, y goes to positive infinity.

You also have to have the graph cross the x-axis three times, at x = 2, x = 3 and x = -6.

Your graph should look like this:

34. A public radio station held a fund-raiser. The table below summarizes the donor category and method of donation.

To the nearest thousandth, find the probability that a randomly selected donor was categorized as a supporter, given that the donation was made online.

Do these data indicate that being a supporter is independent of donating online? Justify your answer.

Answer:

There were 3216 donations made onlie. Of those, 1200 were from supporters. Probability is 1200/3216 = 0.3731..., or 0.373.

There were a total of 1600 supporters out of 4286 donations made, and 1600/4286 = 0.373. So, yes, it is indepenpent because the probability of being a supporter is equal to the probability of being a supporter who donated online.

35. Algebraically solve the system:
(x - 2)² + (y - 3)² = 20
y = -2x + 7

Answer:

The first equation is a circle and the second circle is a line. There may be 0, 1, or 2 possible solutions.

Substitute -2x + 7 for y in the first equation and solve the quadratic equation.

(x - 2)² + (-2x + 7 - 3)² = 20
(x - 2)² + (-2x + 4)² = 20
x² - 4x + 4 + 4x² - 16x + 16 = 20
5x² - 20x + 20 = 20
5x² - 20x = 0
5x(x - 4) = 0
x = 0 or x = 4

Find the matching y-values.

y = -2(0) + 7 = 7, (0,7)
y = -2(4) + 7 = -1, (4,-1)

36. On a certain tropical island, there are currently 500 palm trees and 200 flamingos. Suppose the palm tree population is decreasing at an annual rate of 3% per year and the flamingo population is growing at a continuous rate of 2% per year.

Write two functions, P(x) and F(x), that represent the number of palm trees and flamingos on this island, respectively, x years from now.

State the solution to the equation P(x) = F(x), rounded to the nearest year. Interpret the meaning of this value within the given context.

Answer:

Both functions are exponential. P can be written using the rate of decrease, compounded annually. Since F is continuous, e should be used.

P(x) = 500(1 - .03)^x
F(x) = 200e^.02x

For the second part, you can graph the two functions and find the intersection point. Since it needs to be to the nearest year, you can look at the table of values.

The two will intersect around (18.159, 287.580)

There will be the same number of flamingoes and palm trees in 18 years.

End of Part III

How did you do?

More to come. Comments and questions welcome.

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