To solve for which equation is equivalent to the given formula:
[tex]\[ y = a(x-h)^2 + k \][/tex]
We will go through each option and see if we can transform the given formula into any of the given options.
### Option A:
[tex]\[ k = y + (x-h)^2 \][/tex]
Starting from [tex]\( y = a(x-h)^2 + k \)[/tex]:
1. Subtract [tex]\( a(x-h)^2 \)[/tex] from both sides:
[tex]\[ y - a(x-h)^2 = k \][/tex]
Clearly, this does not match with [tex]\( k = y + (x-h)^2 \)[/tex].
### Option B:
[tex]\[ x = \pm \sqrt{\frac{y-k}{a}} - h \][/tex]
Starting from [tex]\( y = a(x-h)^2 + k \)[/tex]:
1. Subtract [tex]\( k \)[/tex] from both sides:
[tex]\[ y - k = a(x-h)^2 \][/tex]
2. Divide both sides by [tex]\( a \)[/tex]:
[tex]\[ \frac{y - k}{a} = (x-h)^2 \][/tex]
3. Take the square root of both sides:
[tex]\[ \sqrt{\frac{y - k}{a}} = x-h \][/tex]
4. Add [tex]\( h \)[/tex] to both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \pm \sqrt{\frac{y - k}{a}} + h \][/tex]
So, this matches with [tex]\( x = \pm \sqrt{\frac{y-k}{a}} - h \)[/tex].
### Option C:
[tex]\[ h = x - \left(\frac{y-k}{a}\right)^2 \][/tex]
Starting again from [tex]\( y = a(x-h)^2 + k \)[/tex]:
1. Subtract [tex]\( k \)[/tex] from both sides:
[tex]\[ y - k = a(x-h)^2 \][/tex]
2. Divide both sides by [tex]\( a \)[/tex]:
[tex]\[ \frac{y - k}{a} = (x-h)^2 \][/tex]
3. Notice that this step already doesn't match the desired form in Option C, so we can disregard it.
### Option D:
[tex]\[ a = \frac{y - k}{(x-h)^2} \][/tex]
Starting from [tex]\( y = a(x-h)^2 + k \)[/tex]:
1. Subtract [tex]\( k \)[/tex] from both sides:
[tex]\[ y - k = a(x-h)^2 \][/tex]
2. Divide both sides by [tex]\( (x-h)^2 \)[/tex]:
[tex]\[ a = \frac{y - k}{(x-h)^2} \][/tex]
This matches with [tex]\( a = \frac{y - k}{(x-h)^2} \)[/tex].
Thus, we can see that both Option B and Option D are correct transformations of the given formula. However, since we are asked to select the correct answer as per our previous instruction, we choose:
[tex]\[ \boxed{B} \][/tex]
Therefore, the correct answer is:
[tex]\[ x = \pm \sqrt{\frac{y-k}{a}} - h \][/tex]
