how do i solve f(x)=a(x-h)squared +k for minimum wage per year if the year is 1940 and the wage is 0.25

Start by telling us what x, k, a and h mean

you need to know somethign about the variable, and the variable shift h (what units?)

To solve the equation f(x) = a(x-h)^2 + k for the minimum wage per year, given that the year is 1940 and the wage is 0.25, you need to substitute the values of x, a, h, and k into the equation and solve for the unknown.

In this case, we have f(x) = 0.25, a = ?, h = ?, k = 0.25, and x = 1940.

Since we are looking for the minimum wage per year, we can assume that a is a positive value because the square of (x - h) will always be positive or zero.

Now, let's substitute the known values into the equation:

0.25 = a(1940 - h)^2 + 0.25

We'll simplify the equation by subtracting 0.25 from both sides:

0.25 - 0.25 = a(1940 - h)^2

0 = a(1940 - h)^2

Since a is positive, we can divide both sides of the equation by a:

0 / a = (1940 - h)^2

Now we need to solve for (1940 - h)^2. Taking the square root of both sides will help us get rid of the square:

√(0 / a) = √(1940 - h)^2

0 = 1940 - h

Rearranging the equation to solve for h:

h = 1940

Therefore, the value of h is 1940.

To find the value of a, we will use the fact that the wage is 0.25 at x = 1940:

f(1940) = a(1940 - 1940)^2 + 0.25

0.25 = a(0)^2 + 0.25

0.25 = 0a + 0.25

0.25 = 0.25

Since this equation holds true for any value of a, there are infinitely many possible values for a.

Hence, the minimum wage per year for the given equation and data is $0.25 in the year 1940. The value of h is 1940, and the value of a can take any positive value.