To determine whether the function [tex]\( y = x^2 - 4x - 1 \)[/tex] has a minimum or maximum vertex, we can follow these steps:
1. Find the first derivative of the function:
The first derivative, [tex]\( y' \)[/tex], will indicate where the function's slope is zero, showing the critical points (potential maximums, minimums, or points of inflection).
[tex]\[
y' = \frac{d}{dx} (x^2 - 4x - 1) = 2x - 4
\][/tex]
2. Set the first derivative equal to zero to find critical points:
[tex]\[
2x - 4 = 0
\][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[
x = 2
\][/tex]
3. Find the second derivative of the function:
The second derivative, [tex]\( y'' \)[/tex], will help us determine the concavity of the function at the critical points. The nature of the concavity (concave up or concave down) will determine if we have a minimum or maximum vertex.
[tex]\[
y'' = \frac{d}{dx} (2x - 4) = 2
\][/tex]
4. Evaluate the second derivative at the critical point [tex]\( x = 2 \)[/tex]:
[tex]\[
y''(2) = 2
\][/tex]
Since the second derivative is positive ([tex]\( y''(2) > 0 \)[/tex]), the function is concave up at [tex]\( x = 2 \)[/tex]. A positive second derivative indicates that the critical point is a local minimum.
Therefore, the function [tex]\( y = x^2 - 4x - 1 \)[/tex] has a minimum vertex at [tex]\( x = 2 \)[/tex].
