Question

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!

Answer 1

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

Answer 2

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