KarunMendiratta wrote:
\(f(x) = ax^2 + bx + c\)
The equation above represents a parabola in the xy-plane. If \(a < 0\) and the x-intercepts of the parabola are \(-1\) and \(3\), which of the following could be the vertex of the parabola?
A. \((2, 5)\)
B. \((1, -3)\)
C. \((2, 3)\)
D. \((1, 4)\)
E. \((2, -2)\)
Explanation:The vertex of Parabola lies at the mid-point of x-intercepts (or x values for which we have same y values)
Here, \(x_v = \frac{3 + (-1)}{2} = 1\)
Eliminate A, C, and E
Since, \(a < 0\), the Parabola will open downwards with the y-coordinate of vertex (Maximum point in this case) as a +ve value
Hence, option D