Answer:
Step-by-step explanation:
Hi there,
The graph indicated is showing a horizontal asymptote. In fact, it is showing both a horizontal and a <em>vertical </em>asymptote.
To tell which type it is, notice where the graph "shoots off" and almost forms an imaginary straight line in one direction. Using this logic, the horizontal asymptote will be exactly horizontal, parallel to x-axis, and vertical asymptote will be exactly vertical, parallel to y-axis.
With this graph, we notice the horizontal asymptote is at y=0, where the x-axis is. The vertical asymptote is bit more difficult to determine graphically, but can definitely say it is past x=-10. We could determine it if we had the function, but that is not necessary for this question.
Study well, and persevere. If you liked this solution, leave a Thanks or give a rating!
thanks,
Answer:
2x^2+4x-16
Step-by-step explanation:
The quadratic can be written as
f(x) = a(x-z1)(x-z2) where z1 and z2 are the roots
f(x) = a (x-2)(x- -4)
a is the leading coefficient
f(x) = 2(x-2)(x+4)
= 2(x^2 -2x+4x-8)
= 2(x^2 +2x-8)
= 2x^2 +4x-16
Answer:
Polynomial
Step-by-step explanation:
Answer:
you need to provide the figure
