Answer: The vertex of the parabola (quadratic function) is (-2,-4)
Fourth option: (-2,-4)
Solution:
y=x^2+4x
y=ax^2+bx+c; a=1, b=4, c=0
Vertex: V=(h,k)
h=-b/(2a)
h=-4/(2(1))
h=-4/2
h=-2
y=x^2+4x
k=y=h^2+4h
k=(-2)^2+4(-2)
k=4-8
k=-4
Vertex: V=(h,k)
Vertex: V=( -2, -4)
The slope of the line is positive
If you are asking to simplify, this is what I got. Hope I was able to help
In the usual sense, it has no inverse function. Since it is not an injection (f(0)=f(2)=0), an inverse function would have to map 0 both to 0 and 2, and that’s impossible.
However, if you restrict the domain to [1,+oo> (to make it injective), and the codomain to [-1,+oo> (to make it surjective), then it has an inverse function, given by g(y)=1+sqrt(y+1).
I hope you don't find the solution too messy. If there is anything you don't get. Let me know and I will explain it:
