Which parabola has a minimum value at its vertex?

• A. $y=x^2-4x+1$
• B. $y=-2x^2+3x-5$
• C. $y=-x^2+6$
• D. $y=-1x^2-8x+2$

Answer: (A)

Verified step by step guidance

1. Recall that a parabola given by the quadratic function \(y = ax^2 + bx + c\) has a vertex that represents a minimum value if the parabola opens upwards. This happens when the coefficient \(a\) is positive (\(a > 0\)).
2. Identify the coefficient \(a\) in each given quadratic equation to determine whether the parabola opens upwards or downwards:
3. For example, in the equation \(y = x^2 - 4x + 1\), the coefficient \(a\) is 1, which is positive, so this parabola opens upwards and has a minimum at its vertex.
4. In contrast, if \(a\) is negative (like in \(y = -2x^2 + 3x - 5\)), the parabola opens downwards and has a maximum at its vertex, not a minimum.
5. Therefore, to find the parabola with a minimum value at its vertex, select the quadratic equation where \(a > 0\).