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If y = x^3- 3x^2 + 2, find the value of y if a.) x=0, b.) x=2, c.) x=-1, d.) x=a, and e.) x=2x.

User Isgoed
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1 Answer

13 votes
13 votes

Answer:

a.) y = 2

b.) y = -2

c.) y = -2

d.) y = a³ - 3a² + 2

e.) y = 8x³ - 12x² + 2

Explanation:

a.) x = 0


y = x^3- 3x^2 + 2\\y = (0)^3- 3(0)^2 + 2\\y = 0- 0 + 2\\y = 2

b.) x = 2


y = x^3- 3x^2 + 2\\y = (2)^3- 3(2)^2 + 2\\y = 8 - 12 + 2\\y = -2

c.) x = -1


y = x^3- 3x^2 + 2\\y = (-1)^3- 3(-1)^2 + 2\\y = -1 - 3 + 2\\y = -2

d.) x = a


y = x^3- 3x^2 + 2\\y = (a)^3- 3(a)^2 + 2\\y = a^3 - 3a^2 + 2

e.) x = 2x


y = x^3- 3x^2 + 2\\y = (2x)^3- 3(2x)^2 + 2\\y = (8)x^3- 3(4)x^2 + 2\\y = 8x^3 - 12x^2 + 2

User Ola Wiberg
by
2.9k points
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