saswata4s wrote:

If y<0, which of the following must decrease as y decreases?

I. \(10y -100\)

II. \(\frac{y - 1}{y}\)

III. \(y^2 - y\)

(A) I only

(B) III only

(C) I and II only

(D) I and III only

(E) II and III only

Dear

saswata4s,

I'm happy to respond.

This is a great question!

Statement I is relatively easy. As y decreases, any positive multiple of y will decrease, so 10y decreases. If something decreases, adding or subtracting a constant won't change the pattern, so if 10y decreases, then 10 y - 100 will decrease. Statement 1 is a

yes.

Statement III is also relatively straightforward. This is a parabola. Parabolas go up on both sides, so this increase for both positive and negative values of y in the long run. Depending on where the vertex is, it might decrease then increase--we don't need to calculate the vertex. The point is, we KNOW this will not keep decreasing forever and ever. Statement III is a

no.

Statement II is the tricky one. \(\frac{y - 1}{y}\) = 1 - \(\frac{1}{y}\). The one doesn't matter: that's another constant. The real question is what happens to \(-\frac{1}{y}\) when y < 0? Well, as y < 0 decreases, we will get negative numbers with larger and large absolute values. The negative sign (technically, an "

opposite sign" in this case) would make all those negative positives, so this would have the same result as positive y's increasing, starting from just above zero and going to larger and large positive values. We know, in that case, \(\frac{1}{y}\) would continually decrease. Thus, for decreasing negative numbers, \(-\frac{1}{y}\) continually decreases. Statement II is a

yes.

OA =

(C) Great math question!

Does all this make sense?

_________________

Mike McGarry

Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)