Why is 3 the exponent for the factor 2?
2 answers:
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We know that
the equation of a vertical parabola in vertex form
y=a*(x-h)²+k
(h,k)------> (0,5)
y=a*(x-0)²+5
y=a*x²+5
substitute the point (2,9) in the equation
9=a*(2)²+5------> 9=4*a+5-------> 4*a=9-5-----> 4*a=4-----> a=1
the equation of the vertical parabola is
y=x²+5
the equation of a horizontal parabola in vertex form
x=a*(y-k)²+h
(h,k)------> (0,5)
x=a*(y-5)²+0
x=a*(y-5)²
substitute the point (2,9) in the equation
2=a*(9-5)²------> 2=16*a------> a=1/8
the equation of the horizontal parabola is
x=(1/8)*(y-5)²
the answer is
the equation of the vertical parabola is y=x²+5
the equation of the horizontal parabola is x=(1/8)*(y-5)²
see the attached figure
the equation of a vertical parabola in vertex form
y=a*(x-h)²+k
(h,k)------> (0,5)
y=a*(x-0)²+5
y=a*x²+5
substitute the point (2,9) in the equation
9=a*(2)²+5------> 9=4*a+5-------> 4*a=9-5-----> 4*a=4-----> a=1
the equation of the vertical parabola is
y=x²+5
the equation of a horizontal parabola in vertex form
x=a*(y-k)²+h
(h,k)------> (0,5)
x=a*(y-5)²+0
x=a*(y-5)²
substitute the point (2,9) in the equation
2=a*(9-5)²------> 2=16*a------> a=1/8
the equation of the horizontal parabola is
x=(1/8)*(y-5)²
the answer is
the equation of the vertical parabola is y=x²+5
the equation of the horizontal parabola is x=(1/8)*(y-5)²
see the attached figure