Symmetrical Parabola's: opposit "a"

Thanks for the feedback, come back and tell your friends. Sat-tutor is about to have a whole lot more, including practice problems and video explainations.

When two parabolas are defined by quadratic equations that have "opposite" coeffecients, the SAT is giving you a REALLY BIG clue. Looking something like:

f(x) = x^2

f(x) = -x^2 + k (where ^ is the symbol for exponent & k is a constant)

The SAT will give you points where these two intersect. Notice that they are both symmetrical about the Y-axis, making the two points of intersect equal distance from the Y-axis (where x = 0). Let's call those two points: P & Q

They give some clue about these two points, something like: the distance of line segment PQ is 6. What is the value of k?

First step:

The distance between PQ is important. Since the parabola's intersent at a point x, -x, the total distance between them, 2x is equal to 6. 2x = 6. x = 3, -x = -3

Second step:

Substitute x = 3, into the first equation. 3^2 = 9. The points of intersection P & Q are (3, 9) & (-3, 9)

Third step:

Substitute P or Q into the second equation: f(x) = -x^2 + k

9 = -(3)^2 + k

9 = -9 + k

18 = k

The one thing that I highly recommend SAT prep students is to buy "The Official Study Guide" published by The College Board. It has 8 practice tests from REAL SAT's. It is the only one that has REAL practice problems - so it is the best piece of intelligence.

I have my students go through and do every parabola problem in The Guide.

Check back in a few days and I will post all of the parabola problems in The Guide, with notation for "Symmetrical Parabola" problems.

Excuse the really crude drawing but it is now 3:30 in the morning. I had crashed in a chair after coming home from my night school class & was on my way to bed when I saw you had posted, so I cranked out this thought before I forgot it.

Now that I know how easy it is to make a drawing and post it - watch out!