In the xy-plane, the line $y = 5$ intersects the parabola $y = x^2 + bx + 9$ at exactly one point. What is the value of $b$ if $b > 0$?
A$2$
B$3$
C$5$
D$4$
Answer & Solution
Correct answer: D. $4$
Substitute $y = 5$: $5 = x^2 + bx + 9$, so $x^2 + bx + 4 = 0$. Exactly one solution requires discriminant zero: $b^2 - 4(1)(4) = 0$, so $b^2 = 16$, $b = 4$ (taking the positive root).
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