Respuesta :
The graph of the function [tex]f(x) =-x^2-4x + 2[/tex] is parabola with branches going down in the negative direction of y-axis.
The vertex of parabola has coordinates:
[tex]x_v=\dfrac{-(-4)}{2\cdot(-1)}=-2, \\ y_v=-(-2)^2-4\cdot (-2)+2=-4+8+2=6[/tex]
Then you can conclude that all x are possible, that means that the dimain is [tex]x\in (-\infty,\infty)[/tex] and the maximum value of y is at the vertex, then the range is [tex](-\infty,6][/tex].The function is increasing for x<-2 and decreasing for x>-2 (since vertex is the maximum point).
When x=0, y=2.
Hence,
The domain is {x|x ≤ –2} - false.
The range is {y|y ≤ 6} - true.
The function is increasing over the interval (–∞ , –2) - true.
The function is decreasing over the interval (−4, ∞) - false.
The function has a positive y-intercept - true.
The vertex of parabola has coordinates:
[tex]x_v=\dfrac{-(-4)}{2\cdot(-1)}=-2, \\ y_v=-(-2)^2-4\cdot (-2)+2=-4+8+2=6[/tex]
Then you can conclude that all x are possible, that means that the dimain is [tex]x\in (-\infty,\infty)[/tex] and the maximum value of y is at the vertex, then the range is [tex](-\infty,6][/tex].The function is increasing for x<-2 and decreasing for x>-2 (since vertex is the maximum point).
When x=0, y=2.
Hence,
The domain is {x|x ≤ –2} - false.
The range is {y|y ≤ 6} - true.
The function is increasing over the interval (–∞ , –2) - true.
The function is decreasing over the interval (−4, ∞) - false.
The function has a positive y-intercept - true.
Answer:
The range is {y|y ≤ 6}
The function is increasing over the interval (–∞ , –2)
The function has a positive y-intercept