Answer:
idk
Step-by-step explanation:
im really stupid I need the points to ask my own question
All irrational numbers are real numbers. ☺
Answer:
Option A
Step-by-step explanation:
Here is how to approach the problem:
We see that all our restrictions for all four answer choices are relatively the same with a couple of changes here and there.
One way to eliminate choices would be to look at which restrictions don't match the graph.
At x<-5, there is a linear function that does have a -2 slope and will intersect the x axis at -7. The line ends with an open circle, so any answer choice with a linear restriction of x less than or equal to -5 is wrong. This cancels out choices C and D.
Now we have two choices left.
For the quadratic in the middle, the vertex is at (-2,6) and the vertex is a maximum, meaning our graph needs to have a negative sign in front of the highest degree term. In our case, none of our quadratics left are in standard form, and instead are in vertex form.
Vertext form is f(x) = a(x-h)^2 + k.
h being the x-coordinate of the vertex and k being the y-coordinate.
We know that the opposite of h will be the actual x-coordinate of the vertex, so if our vertex is -2, we will see x+2 inside the parenthesis. This leaves option A as the only correct choice.
