Question

Given a quadratic function, f(x) = ax 2 + bx + c has a negative leading coefficient and the vertex that has a negative y-coordinate. Determine the number of real zeros of the function.2 real zeros1 real zero1 real zero and 1 imaginary zero2 imaginary zeros

Answer 1

The quadratic function with a negative leading coefficient and a vertex with a negative y-coordinate will have 2 real zeros as it intersects the x-axis at two distinct points.

Step-by-step explanation:

Given a quadratic function, f(x) = ax^2 + bx + c has a negative leading coefficient and the vertex has a negative y-coordinate, the graph of the function opens downwards and the vertex is a maximum point. The number of real zeros of the function can be determined by analyzing the graph of the quadratic function. Since the vertex is below the x-axis and the parabola opens downwards, it will intersect the x-axis at two distinct points, indicating that there are 2 real zeros. The quadratic formula, which is x = (-b ± √(b^2 - 4ac)) / (2a), can also reveal the number of real zeros by evaluating the discriminant (b^2 - 4ac). If the discriminant is positive, there are 2 real zeros; a negative discriminant indicates 2 imaginary zeros; and a zero discriminant means there is exactly 1 real zero (a repeated root).

Answer 2

First, for clarity, please use " ^ " to denote exponentiation: f(x) = ax^2 + bx + c.

Given: a<0
vertex has a negative y-coordinate => f(x) is negative at the vertex.

What can we learn about the discriminant here? Consider b^2 - 4(a)(c).

b^2 is always positive. If a is <0 (as we were told that it is), and if c>0, then b^2 - 4ac will be positive, and because of that, the TWO roots will be real, whether equal or unequal.

However, if c<0, that conclusion no longer holds; you'd have two complex roots (or two imaginary roots).