Step 1 — Extract info from the original graph.
The original equation opens downward and has vertex with .
This tells us .

Step 2 — Identify the vertex of the new equation.
is in vertex form , where vertex .
Rewrite: , so and .
Vertex → eliminates A and B.

Step 3 — Determine the direction of opening.
The leading coefficient is . Since , we have → parabola opens upward → eliminates D.

Answer: C vertex , opens upward.

Pattern: On any vertex-form question : (1) vertex is always — watch the sign carefully; (2) the sign of determines direction, not its magnitude. If involves a negated parameter like , always ask what sign that parameter carries from the given context.

Answer: C

Key insight: is just a horizontal line at height . The number of real solutions equals the number of times that line intersects the graph.

For a cubic with one local max and one local min, a horizontal line crosses it three times only if it passes strictly between the two bumps — i.e., must be strictly between the local min value and the local max value.

From the graph, the local max is near and the local min is near . So must satisfy . The only answer in that range is — as shown by the orange line hitting the curve at exactly 3 points.

Time-saving rule: Never try to solve a cubic algebraically when a graph is given. Just ask: for what -values does a horizontal line hit the curve exactly 3 times? Then scan the answer choices.

Answer: A

Since the leading coefficient is (negative), the parabola opens downward. For it to intersect the -axis at two points, the vertex must be above the -axis — which is satisfied since the vertex -value is .

is the -intercept. Using vertex form:

At :

Since , we get . The only answer choice less than is .

No need to find — just knowing is enough to eliminate B, C, and D instantly.

Pattern: When the SAT gives you a vertex and asks for a coefficient, write vertex form first, plug in the relevant point ( for ), and read off the constraint. A quick sketch makes the reasoning obvious and prevents you from trying to expand anything.

Answer: B

An isosceles right triangle (45-45-90) has two equal legs and hypotenuse . So:

Setting equal to the given perimeter and rationalizing:


The numerator:

Why A is a trap: seeing in the perimeter makes feel like the obvious half — but that ignores the hypotenuse entirely. Always write out before touching the numbers.

Answer: B

Step 1 — Find W.
is on the unit circle, so . With : .

Step 2 — Find the base angle.
is the center. Ray points to = rad. Ray points to .

Key convention: in the standard coordinate system, positive angles are always counterclockwise. Clockwise = negative angle. This is a fixed mathematical convention, the same one behind the unit circle, polar coordinates, and complex numbers.

Going counterclockwise from , you pass through , , and finally reach :

Going clockwise (the short way) would be , but that is a negative rotation: — ruled out by the word "positive" in the question.

Step 3 — Add full rotations.
Any coterminal angle is also valid. For :

Why the other choices fail:

• A) is coterminal with → lands at , not
• C) is coterminal with → lands at
• D) is coterminal with → lands at

Trigger words: whenever you see "positive measure" on a unit circle problem, that means counterclockwise only. Find the base counterclockwise angle first, then add and match an answer choice.

Fast elimination trick — remainder mod 4:
All answer choices of the form reduce to one of the four cardinal points based on :

On the FX-991CW (no MOD button): divide by → the decimal part maps directly to the remainder: , , , .

• A) → remainder → ✗
• B) → remainder → ✓
• C) → remainder → ✗
• D) → remainder → ✗

Answer: B

The standard form of a circle is , with center . To get there, complete the square on both and .

Step 1 — Complete the square for .
Coefficient of is . Take half: . Square it: . Add to both sides:

Step 2 — Complete the square for .
Coefficient of is . Take half: . Square it: . Add to both sides:

Step 3 — Read off the center and radius.
Center , radius .

The method — completing the square:
For any term , add to complete the square:

Whatever you add to the left side, you must add to the right side too.

Why the traps exist:

• A) : raw coefficients, not halved — the most common mistake
• C) : correct magnitude but wrong sign — , so
• D) : both errors combined

SAT pattern: Any circle equation not in standard form will require completing the square. It shows up in ~1 question per test. The process is always the same: group terms, group terms, complete the square on each, balance both sides.

Answer: B is impossible.

Setup: When a transversal cuts two parallel lines, all 4 acute angles are equal and all 4 obtuse angles are equal, and they are supplementary:

All valid combinations of 4 angles chosen from the 8:

The only expression with is . Option B gives — a constant that never appears. B is impossible.

Key facts to memorize:

• All 4 acute angles are equal; all 4 obtuse angles are equal
• Acute + obtuse (supplementary, not complementary)
• 2 acute + 2 obtuse always, regardless of — that’s why C is always valid

"NOT" trap: The question asks which one cannot be . Slow down and build the full table before committing to an answer — don’t stop as soon as you find one that works.

Answer:

The line simplifies to , a horizontal line. A horizontal line is tangent to a downward parabola at exactly one point only at the vertex. So the vertex -coordinate must equal .

Method 1 — Vertex -coordinate formula:
For , the vertex -coordinate is:

Here , :