Question

Please help me if you know how to do this!

Answer 1

Question 21

Let's complete the square

y = 3x^2 + 6x + 5

y-5 = 3x^2 + 6x

y - 5 = 3(x^2 + 2x)

y - 5 = 3(x^2 + 2x + 1 - 1)

y - 5 = 3(x^2+2x+1) - 3

y - 5 = 3(x+1)^2 - 3

y = 3(x+1)^2 - 3 + 5

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

Answer: Choice D

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Question 22

Through trial and error you should find that choice D is the answer

Basically you plug in each of the given answer choices and see which results in a true statement.

For instance, with choice A we have

y < -4(x+1)^2 - 3

-7 < -4(0+1)^2 - 3

-7 < -7

which is false, so we eliminate choice A

Choice D is the answer because

y < -4(x+1)^2 - 3

-9 < -4(-2+1)^2 - 3

-9 < -7

which is true since -9 is to the left of -7 on the number line.

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Question 25

Answer: Choice B

Explanation:

The quantity (x-4)^2 is always positive regardless of what you pick for x. This is because we are squaring the (x-4). Squaring a negative leads to a positive. Eg: (-4)^2 = 16

Adding on a positive to (x-4)^2 makes the result even more positive. Therefore (x-4)^2 + 1 > 0 is true for any real number x.

Visually this means all solutions of y > (x-4)^2 + 1 reside in quadrants 1 and 2, which are above the x axis.