Answer:
The real zeros of the quadratic function f(x) = -4x^2 - 6x + 1 are approximately -0.15 and -1.35.
Explanation:
To find the real zeros of the quadratic function f(x) = -4x^2 - 6x + 1, we need to find the values of x that make f(x) equal to zero. We can do this by using the quadratic formula:
x = [-b ± sqrt(b^2 - 4ac)] / 2a
where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c.
In this case, a = -4, b = -6, and c = 1. Substituting these values into the quadratic formula, we get:
x = [-(-6) ± sqrt((-6)^2 - 4(-4)(1))] / 2(-4)
x = [6 ± sqrt(52)] / (-8)
x = [6 ± 2sqrt(13)] / (-8)
These are the two solutions for the quadratic equation, which we can simplify as follows:
x = (3 ± sqrt(13)) / (-4)
Therefore, the real zeros of the quadratic function f(x) = -4x^2 - 6x + 1 are approximately -0.15 and -1.35.