Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, June 2011)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Algebra 2/Trigonometry Regents, June 2011

Part II: Each correct answer will receive 2 credits. Partial credit is available.


32. If f(x) = x² − 6, find f⁻¹(x).

Answer:

Replace x with y, and replace f(x) with x. Then solve for y.

x = y² − 6

x + 6 = y²

y = + √(x + 6)

f⁻¹(x) = + √(x + 6)

33. Factor the expression 12t⁸ − 75t⁴ completely.

Answer:

Factor completely means that there will most likley be more than one step. You have a common factor is the coefficients, the same variable in both terms and a difference of squares to start with.

12t⁸ − 75t⁴

3t⁴(4t⁴ − 25)

3t⁴(2t² + 5)(2t² - 5)

You could also factor the last binomial again using √(2) and √(5), but that isn't necessary since 2 and 5 are not perfect squares.

34. Simplify the expression (3x^-4y⁵) / (2x³y^-7)^-2 and write the answer using only positive exponents.

Answer:

Since I'm typing this, I'm going to use the Rules for Exponents rather than playing with fractions:

(3x^-4y⁵) / (2x³y^-7)^-2

(3x^-4y⁵) / (4^-1x^-6y¹⁴)

(3x^{(-4 - -6)}y^{(5 - 14)}) / (4^-1)

(3x²y^-9) / (4^-1)

(3*4)x² / y⁹

12x² / y⁹

35. If f(x) = x² − 6 and g(x) = 2^x − 1, determine the value of (g ° f)(−3).

Answer:

Remember that (g ° f)(−3) means g(f(-3)), so you need to find f(-3) first.

f(-3) = (-3)² - 6 = 9 - 6 = 3

g(3) = 2³ - 1 = 8 - 1 = 7

(g ° f)(−3) = 7

More to come. Comments and questions welcome.

More Regents problems.

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