Question

On a coordinate plane, a parabola opens up. It goes through (negative 1, 0), has a vertex at (1, negative 4), and goes through (3, 0).
Which describes all of the values for which the graph is positive and decreasing?

all real values of x where x < −1
all real values of x where x < 1
all real values of x where 1 < x < 3
all real values of x where x > 3

Answer 1

Answer:

All real values of x where x < −1.

Step-by-step explanation:

There is a turning point at x= 1 and f(x) = -4 and it open upwards.

So it decreases for all real values of x  where x < 1. But it passes through the point (-1,  0) after which the values of the function become negative.

So the answer is all real values where x < -1.

Answer 2

Answer:

Option A.

Step-by-step explanation:

It is given that an open up parabola goes through (-1, 0), has a vertex at (1, -4), and goes through (3, 0).

x-intercepts of the function are (-1,0) and (3,0).

The vertex of an upward parabola is the point of minima. The function is decreasing on the left side of the vertex and the function is increasing on the right side of the vertex. It means

Function is positive and decreasing where x<-1

Function is negative and decreasing where -1 < x < 1.

Function is negative and increasing where 1 < x < 3.

Function is positive and increasing where x>3.

Therefore, the correct option is A.