What is the standard form of;
y=√2x-x^2
I assume you mean
y =√(2x-x^2)
so,
y^2 = 2x-x^2
x^2-2x + y^2 = 0
x^2-2x+1 + y^2 = 0+1
(x-1)^2 + y^2 = 1
I'm so sorry i've made a typo wht I mean is y=√25-x^2
y = sqrt(25-x^2).
y^2 = 25 - x^2,
x^2 + y^2 = 25.
To find the standard form of the equation y = √2x - x^2, we need to simplify and rearrange the equation.
Step 1: Replace the square root (√) with a fractional exponent (1/2) to make the equation easier to work with. So, we have y = (2x)^(1/2) - x^2.
Step 2: Square both sides of the equation to eliminate the fractional exponent, resulting in y^2 = (2x) - x^4.
Step 3: Rearrange the terms to get all the terms on one side of the equation and set it equal to zero:
y^2 + x^4 - 2x = 0.
Now we have the equation in the standard form: y^2 + x^4 - 2x = 0.