write in transformational form

Question

y = 1/3x^2-5x+3.

Answer 1

The equation y  = 1/3 x ^2 - 5x + 3

Standard form of parabola is y = ax ^2 + bx + c.

Vertex form is y  = a  (x - h )^2 + k

 y  = 1/3 x ^2 - 5x + 3

Here x² coefficient is 1/3, for perfect square make x² coefficient 1 by multiplying each side by 3.

3y  = x ^2 - 15x + 9

To change the expression into a perfect square trinomial add (half the x coefficient)² to each side

of the expression.

x  coefficient = -15 then (half the x coefficient)² is 225/4.

So add 225/4 to eachside.

3y  + 225/4 = x ^2 - 15x  + 9 + 225/4

3y  + 225/4 = x ^2 - 15x  + 225/4 + 9

3y  + 225/4 = (x - 15/2)^2 + 9

3y  = (x - 15/2)^2 + 9 - 225/4

3y  = (x - 15/2)^2 - 189/4

y  = 1/3(x - 15/2)^2 - 189/12

y  = 0.33(x - 7.5)^2 - 15.75

Now it is in vartex form y  = a (x - h )^2 + k

Transformational form of the parabola y  = 0.33(x - 7.5)^2 +(- 15.75).