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Vertex

If the vertex...

Question

If the vertex of the curve $y = - 2 x^{2} - 4 a x - k$ is $(- 2, 7),$ then the value of $k$ is

1.0

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Solution

Comparing
$y = - 2 x^{2} - 4 a x - k$ with $y = p x^{2} + q x + r$
We get
$p = - 2, q = - 4 a, r = - k$
So, the coordinates of the vertex is
$∴ (- \frac{q}{2 p}, - \frac{D}{4 p}) \equiv (- 2, 7) \Rightarrow - \frac{(- 4 a)}{- 4} = - 2, - \frac{(16 a^{2} - 8 k)}{- 8} = 7 \Rightarrow a = 2 \Rightarrow - \frac{(16 a^{2} - 8 k)}{- 8} = 7 \Rightarrow 16 \times 4 - 8 k = 56 ∴ k = 1$

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