Math SAT question

[Jessica]

http://www.collegeboard.com/prod_images/math/art-MRO_QE_14.gif
y = (–x)^2 + 1
y = –x^2 + 1
y = |x^2 + 1|
y = |x^2 – 1|
y = |(x – 1)^2|
I think it's 3; because 1 and 2 the x is negative, but the graph is facing up. and it can't be 4 or 5 because the y-int. is 1, not -1.

[Fazor]

y = (–x)^2 + 1
y = –x^2 + 1
y = |x^2 + 1|
y = |x^2 – 1|
y = |(x – 1)^2|

I think it's 3; because 1 and 2 the x is negative, but the graph is facing up. and it can't be 4 or 5 because the y-int. is 1, not -1.

Jeeze! You're good at making me realize just how little of this stuff I remember!

As far as I can tell, you can't rule 1 or 2 out because of the negative x, because the 'squared' would still leave y as a positive number. You can, however, rule them out because the lowest possible value for y would be 1 (because of the + 1), and you can see from the chart that twice, the value of y reaches 0.

The same would hold true for answer #3. I believe the correct answer is #4, because at x = 0 then y = |0^2 - 1| which means y = |-1| leaving y = 1. That is correct for the graph.

But as you move away from zero, say, x = .5, you get:
y = |.5^2 - 1| which is y = |.25 - 1| or y = |.75| or y = .75

So as x starts to move away from zero, y initially continues to drop. At x = 1, you're left with y = 0. Which matches the graph. Then as you continue past x = 1 ( x = 2, 3, etc), the value of y starts to raise again.

[bobjenkins]

Jeeze! You're good at making me realize just how little of this stuff I remember! sí, no recuerdo nada de la matemáticas:lol:

[brute]

Jeeze! You're good at making me realize just how little of this stuff I remember!

As far as I can tell, you can't rule 1 or 2 out because of the negative x, because the 'squared' would still leave y as a positive number. You can, however, rule them out because the lowest possible value for y would be 1 (because of the + 1), and you can see from the chart that twice, the value of y reaches 0.

The same would hold true for answer #3. I believe the correct answer is #4, because at x = 0 then y = |0^2 - 1| which means y = |-1| leaving y = 1. That is correct for the graph.

But as you move away from zero, say, x = .5, you get:
y = |.5^2 - 1| which is y = |.25 - 1| or y = |.75| or y = .75

So as x starts to move away from zero, y initially continues to drop. At x = 1, you're left with y = 0. Which matches the graph. Then as you continue past x = 1 ( x = 2, 3, etc), the value of y starts to raise again.
Yes I agree. This is a good analysis. Without the modulus sign, only the graph of equation 4 would drop below the x axis. The modulus sign flips this part of the graph over, making it positive. Answer 4

[Jessica]

Thanks guys! ^_^

[Fazor]

Since I more or less just worked through the problem in my own, rambling style (my personal approach); let me add one tip:

With these kinds of problems, it often helps to start by solving for x = 0 and then y = 0.