To find the minimum value of the parabola given by the equation \( y = x^2 - 6x \), we can complete the square.
-
Start with the original equation: \[ y = x^2 - 6x \]
-
To complete the square, we need to take the coefficient of \( x \) (which is -6), divide it by 2, and square it: \[ \left(-\frac{6}{2}\right)^2 = (-3)^2 = 9 \]
-
Add and subtract this square (9) inside the equation: \[ y = (x^2 - 6x + 9) - 9 \]
-
Now, rewrite the equation: \[ y = (x - 3)^2 - 9 \]
Now we have the equation in vertex form \( y = (x - h)^2 + k \), where \((h, k)\) is the vertex of the parabola. Here, the vertex is at \((3, -9)\).
Therefore, the minimum value of the parabola (since it opens upwards) occurs at the vertex \( y = -9 \).
Thus, the minimum value of the parabola \( y = x^2 - 6x \) is: \[ \boxed{-9} \]
5 answers