SAT Math: Line y = -4x + k Intersects Parabola y = 2x² + 8x - 5 at One Point
Set 2x^2 + 8x - 5 = -4x + k => 2x^2 + 12x - (5 + k) = 0. For exactly 1 intersection, Delta = 12^2 - 4(2)(-5 - k) = 0 => 144 + 40 + 8k = 0 => 184 + 8k = 0 => k = -23.
Exact Question Text
In the $xy$-plane, the line $y = -4x + k$ intersects the parabola $y = 2x^2 + 8x - 5$ at exactly one point, where $k$ is a constant. What is the value of $k$?
Step-by-Step Solution
- Equate the functions:
$$ 2x^2 + 8x - 5 = -4x + k \implies 2x^2 + 12x - (5 + k) = 0 $$
- Set Discriminant to Zero:
$$ (12)^2 - 4(2)(-5 - k) = 0 $$
$$ 144 + 40 + 8k = 0 $$
$$ 184 + 8k = 0 \implies k = \mathbf{-23} $$
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