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What is the equation of a parabola whose vertex is (-6,0) and whose graph is half as...

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kristenmarieb... | Student, Grade 10 | (Level 1) Valedictorian

Posted June 16, 2013 at 1:02 AM via web

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What is the equation of a parabola whose vertex is (-6,0) and whose graph is half as tall as y = x^2 and opens upside down? 

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llltkl | College Teacher | (Level 3) Valedictorian

Posted June 16, 2013 at 2:12 AM (Answer #1)

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The equation of a parabola whose vertex is at (h,k) is given by

 y = a(x - h)² + k

So many parabolas have the vertex at (-6,0) and

open down.  We'll find one.

  Substitute (h,k) = (-6,0) in the standard equation

     y = a(x - h)^2 + k

     y = a(x - (-6) )^2 + 0    

     y = a(x + 6)^2 = ax2 +12ax+36a

But then choose any negative number for a, and its

graph will open downward.  For instance, letting

a = -1 gives this parabola:

    y = -(x + 6)^2

As for the height of the parabola y = x^2, it passes through the origin and opens upward. The question of tallness of the parabola is thus meaningless.

Assuming its width to be double that of the parabola y = x^2 (thus wider and shorter in appearance), the equation then takes the form:

y = -2(x + 6)^2

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