Thursday, January 20, 2022

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) = 2x − 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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