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Question 815009: Find the vertex, focus, and directrix of the parabola
Y^2+14y+4x+45=0
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! y^2 + 14y + 4x + 45 = 0
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-y^2 - 14y - 45 = 4x
x = -0.25y^2 - 3.5y - 11.25
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the above quadratic equation is in standard form, with a=-0.25, b=-3.5, and c=-11.25
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to solve the quadratic equation, plug this:
-0.25 -3.5 -11.25
into this: https://sooeet.com/math/quadratic-equation-solver.php
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note:
the above solver is for parabolas with vertical axes of symmetry:
Y(x) = ax^2 + bx + c
the problem parabola has a horizontal axis of symmetry:
x = ay^2 + by + c
so just flip x and y in the solver's results, as done below:
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vertex is a maximum at ( 1, -7 )
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focus is ( 0, -7 )
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directrix is x = 2
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