Question

Which of the following equations is of a parabola with a vertex at (0, 6)?

y = (x - 6)2
y = (x + 6)2
y = x2 - 6
y = x2 + 6

Answer 1

The answer is:

$y = x^{2} + 6$
EXPLANATION

I dunno, this is just a technique. I believe you know that the vertex form of a quadratic equation is

$y = a(x - h)^{2} + k$
Referring to the given vertex, you have (0,6). So, h should be 0 and k should be 6.

You can see that h is absent in the last two equations given that it is only x^2 and not x minus h.

Going to k, we can find that only the third equation has fulfilled a positive k (the fourth has -6).

In problems like this, I think you can already manage that easy to choose which among the quadratic equations show a particular vertex.

Answer 2

Answer:

The equation of a parabola with a vertex at (0, 6) is:

                              $y=x^2+6$

Step-by-step explanation:

We know that the general equation of a upward or a downward parabola is given by:

                      $y=a(x-h)^2+k$

where if a>0 then it is a upward open parabola.

and if a<0 then it is a downward open parabola.

Also, we have vertex at (h,k)

Now we are given that the vertex is at (0,6)

and also by looking at the options we see that a=1

Hence we have:

h=0 and k=6

Hence, the equation of the parabola is:

                            $y=x^2+6$