Hello. I just wanna ask if how I could simply find the eq'n of parabola satisfying the ff. properties:
Passing through (-3,6); V(0,0); axis of the parabola is the y-axis.
Thanks!
Equation of quadratic parabola:
y = a x² + b x + c
Since for x = 0 , y = 0
0 = a ∙ 0² + b ∙ 0 + c
0 = c
c = 0
x coordinate of vertex:
x = - b / 2 a
In this case, x coordinate of vertex is 0
0 = - b / 2 a
Multiply both sides by - 2 a
0 = b
b = 0
The equation becomes:
y = a x²
Put x = - 3 , y = 6 in this equation
6 = a ( - 3 )²
6 = 9 a
Divide both sides by 9
6 / 9 = a
3 ∙ 2 / 3 ∙ 3 = a
2 / 3 = a
a = 2 / 3
Equation of your quadratic parabola:
y = 2 / 3 x²
For a parabola with vertex (0,0) and a vertical axis of symmetry, the equation must be
y = a x^2
but (-3,6) lies on it, so
6 = a(-3)^2
a = 6/9 = 2/3
equation: y = (2/3)x^2