Splendidgirl666 wrote:
What is the minimum value of the function y = x^2 - 4x - 5?
I would just find the vertex of the parabola that corresponds with the graph of the quadratic function.
The y-coordinate of the vertex is the minimum value of a parabola that, as here, opens upward (positive coefficient for \(a\), squared term is x).
I do not like completing squares in order to put the quadratic function in vertex form. Technically, and especially when the function is in vertex form, the vertex is (h,k). I'll use x and y here.
Just two steps: 1) find the axis of symmetry and x-coordinate of vertex, then 2) plug that value into the function to find y-coordinate, which is this function's minimum value.
Use \(\frac{-b}{2a}\) to find the axis of symmetry and x-coordinate of the vertex.
\(x =\frac{-b}{2a} =\frac{-(-4)}{2} = 2\)
\(x\) = 2
Evaluate y at x (plug x = 2 into the function) to find y-coordinate:
\(y = x^2 - 4x - 5\)
\(y = 2^2 - 4(2) - 5\)
\(y = -9\)
The minimum value of an upward curving parabola is the y-coordinate of the vertex.
Hence the minimum value of the function \(y = x^2 - 4x - 5\) is -9. y will never be less than -9.
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