Bluebook Test 4 Math Module 2 Question 20: y = 3x² - bx + 12 One x-Intercept
For y = 3x^2 - bx + 12 to intersect the x-axis at exactly one point, the discriminant must be zero: (-b)^2 - 4(3)(12) = 0 => b^2 - 144 = 0 => b = 12 (since b > 0). Correct Option: B (12).
Exact Question Text
In the $xy$-plane, the graph of the equation $y = 3x^2 - bx + 12$, where $b$ is a constant, intersects the $x$-axis at exactly one point. If $b > 0$, what is the value of $b$?
- A) 6
- B) 12
- C) 24
- D) 144
Step-by-Step Solution
- Condition for 1 intersection: $\Delta = b^2 - 4ac = 0$.
- $(-b)^2 - 4(3)(12) = 0 \implies b^2 - 144 = 0 \implies b = 12$ (given $b > 0$).
- Correct Answer: B (12)
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