What are the vertex, focus and directrix of $y = 2 x^{2} + 11 x - 6$ $y=2x^2 +11x-6$ ?

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What are the vertex, focus and directrix of $y = 2 x^{2} + 11 x - 6$ $y=2x^2 +11x-6$ ?

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Answer 1 · 2016-12-19T07:51:01

The vertex is $= (- \frac{11}{4}, - \frac{169}{8})$ $=(-11/4,-169/8)$
The focus is $= (- \frac{11}{4}, - \frac{168}{8})$ $=(-11/4,-168/8)$
The directrix is $y = - \frac{170}{8}$ $y=-170/8$

Explanation:

Let rewrite the equation

$y = 2 x^{2} + 11 x - 6$ $y=2x^2+11x-6$

$= 2 (x^{2} + \frac{11}{2} x) - 6$ $=2(x^2+11/2x)-6$

$= 2 (x^{2} + \frac{11}{2} x + \frac{121}{16}) - 6 - \frac{121}{8}$ $=2(x^2+11/2x+121/16)-6-121/8$

$y = 2 {(x + \frac{11}{4})}^{2} - \frac{169}{8}$ $y=2(x+11/4)^2-169/8$

$y + \frac{169}{8} = 2 {(x + \frac{11}{4})}^{2}$ $y+169/8=2(x+11/4)^2$

$\frac{1}{2} (y + \frac{169}{8}) = {(x + \frac{11}{4})}^{2}$ $1/2(y+169/8)=(x+11/4)^2$

This is the equation of the parabola

${(x - a)}^{2} = 2 p (y - b)$ $(x-a)^2=2p(y-b)$

The vertex is $= (a, b) = (- \frac{11}{4}, - \frac{169}{8})$ $=(a,b)=(-11/4,-169/8)$

The focus is $= (a, b + \frac{p}{2}) = (- \frac{11}{4}, - \frac{169}{8} + \frac{1}{8})$ $=(a,b+p/2)=(-11/4,-169/8+1/8)$

$= (- \frac{11}{4}, - \frac{168}{8})$ $=(-11/4,-168/8)$

The directrix is $y = b - \frac{p}{2}$ $y=b-p/2$

$\Rightarrow$ $=>$ , $y = - \frac{169}{8} - \frac{1}{8} = - \frac{170}{8}$ $y=-169/8-1/8=-170/8$

graph{(y-2x^2-11x+6)(y+170/8)=0 [-14.77, 10.54, -21.49, -8.83]}

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What are the vertex, focus and directrix of $y = 2 x^{2} + 11 x - 6$ $y=2x^2 +11x-6$ ?

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Explanation:

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What are the vertex, focus and directrix of $y = 2 x^{2} + 11 x - 6$ $y=2x^2 +11x-6$ ?