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The equation of a parabola is given. y=−112x2−2x−1 What are the coordinates of the focus of the parabol

Respuesta :

IF the equation is y=−1/12x^2−2x−1

then -12y=x^2+24x+12

-12y+132=x^2+24x+144

-12(y-11)=(x+12)^2

so the vertex of the parabola is at (-12, 11)

semi axis length=12/4=3

the focus is at (-12, 11-3) = (-12, 8)
using a graphing calculator, u can plot y=−112x2−2x−1

it is a concave parabola with focus at approx. (-0.0089, -0.9933)

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