Respuesta :
Using it's vertex, it is found that the range of the quadratic function is given by: [tex][-4.5, \infty)[/tex].
What is the vertex of a quadratic equation?
A quadratic equation is modeled by:
[tex]y = ax^2 + bx + c[/tex]
The vertex is given by:
[tex](x_v, y_v)[/tex]
In which:
[tex]x_v = -\frac{b}{2a}[/tex]
[tex]y_v = -\frac{b^2 - 4ac}{4a}[/tex]
Considering the coefficient a, we have that:
- If a < 0, the vertex is a maximum point, and the range is [tex](-\infty, y_v][/tex].
- If a > 0, the vertex is a minimum point, and the range is [tex][y_v, \infty)[/tex].
In this problem, the function is:
h(x) = 2x² - 2x - 4.
Hence the coefficients are a = 2, b = -2, c = -4, and the minimum value of the range is given by:
[tex]y_v = -\frac{(-2)^2 - 4(2)(-4)}{4(2)} = -4.5[/tex]
Hence the range is [tex][-4.5, \infty)[/tex].
More can be learned about quadratic functions at https://brainly.com/question/24737967
