What's the minimum y-value on the graph of y= f(x)

Question

f(x) = x^2 - 4x - 2

the minimum y-value is y=?

Answer 1

You can get it by completing the square.

f(x) = x^2 - 4x - 2 = x^2-4x+4 - 6 = (x-2)^2 - 6

Since (x-2)^2 ≥ 0, and f(x) is the sum of (x-2)^2 and -6, when x-2, you get the minimum of f(x).

So, fmin = -6.

Answer 2

The minimum value is at the vertex, so all you need to do is find the y-coordinate of the vertex. Here are the steps to find the vertex from your standard form equation:

 

The equation you have is in standard form (y = ax^2 + bx + c), so we can use the formula -b/(2a) to find the x-coordinate of the vertex.

x = -(-4)/(2*1)

x = 2

 

Now plug that value into your equation for x to solve for the y-coordinate of the vertex.

y = 2^2 -4(2) -2

y = -6

 

The minimum y-value is at -6.

You could also solve this by graphing it and looking for the vertex.