How do I solve for x in:
y = -(x-4)^2 + 4
This is not calculus; it is algebra.
That can be rearranged to give you
(x-4)^2 = 4-y
Then take the square root of both sides and add 4.
x = sqrt(4-y)+4
lol yeah algebra, ok that's what I got, just making sure.
To solve for x in the equation y = -(x-4)^2 + 4, you can follow these steps:
Step 1: Rewrite the equation:
Start by isolating the squared term on one side of the equation. Let's subtract 4 from both sides:
y - 4 = -(x - 4)^2
Step 2: Remove the negative sign:
To eliminate the negative sign, you can multiply both sides of the equation by -1:
-(y - 4) = (x - 4)^2
Now, the equation can be rewritten without the negative sign:
4 - y = (x - 4)^2
Step 3: Take the square root of both sides:
To solve for x, you need to get rid of the square on the right side of the equation. The square root is the inverse operation of squaring, so taking the square root of both sides cancels out the square term:
√(4 - y) = x - 4
Step 4: Add 4 to both sides:
Now, add 4 to both sides of the equation to isolate x:
√(4 - y) + 4 = x
Thus, the solution for x is x = √(4 - y) + 4.