SOLUTION: Find the equation of the parabola that passes through (0,0) (6,0) (3,27)
y=0; when x=6; and y=0 when x=0.
-27=k(x-6)(x-0)
k=9/2
=9/2(x^2-6x)
=9/2x^2-27
However, the a
Algebra ->
Quadratic Equations and Parabolas
-> SOLUTION: Find the equation of the parabola that passes through (0,0) (6,0) (3,27)
y=0; when x=6; and y=0 when x=0.
-27=k(x-6)(x-0)
k=9/2
=9/2(x^2-6x)
=9/2x^2-27
However, the a
Log On
Question 932492: Find the equation of the parabola that passes through (0,0) (6,0) (3,27)
y=0; when x=6; and y=0 when x=0.
-27=k(x-6)(x-0)
k=9/2
=9/2(x^2-6x)
=9/2x^2-27
However, the answer given is 3x^2-18x
Where have I gone wrong?
You can put this solution on YOUR website! (0,0) (6,0) (3,-27)* Assuming it is '-27'
y = a(x-6)(x-0) Good Work
y = a(x^2 - 6x)
...
P(3,-27)
-27 = a(-9)
3 = a
..........
y = 3(x^2-6x)
y = 3x^2 - 18x
