Help I'm going to fail

Question

Write an equation for a parabola that is congruent to y=1/3x^2, has vertex (3,-5)

Answer 1

Ashley, it seems we are moving our parabola right 3 and down 5. How do we do that??

 

Moving horizontally goes with x, but opposite the way we want it to go, so x becomes x - 3.

 

Moving vertically doesn't go with x. It's just at the end of the equation.

 

y = (1/3)(x - 3)² - 5

Answer 2

I believe y = 1/3 x² is considered the parent function.  To find a congruent parabola, you would need to use vertex form

 

y = a(x-h)² + k where (h, k) are the vertex and a is the constant factor. 

 

Substituting the values in the equation would give you y = -1/3 (x - 3)² + (-5) 

 

So y = -1/3 (x-3)² - 5 would be a congruent parabola.