Which is a possible turning point for the continuous function f(x)? (–2, 0) (0, –2) (2, –1) (4, 0)

We are given coordinates of a continuous function f(x)

(–2, 0)

(0, –2)

(2, –1)

(4, 0).

We need to find the possible turning point for the continuous function.

Note: Turning point is a point on the graph where slope of the curve changes from negative to positive or positive to negative.

A turning point is always lowest or highest point of the curve (where bump of the graph seen).

For the given coordinates we can see that  (–2, 0) and (4, 0) coordinates are in a same line, that is on the x-axis.

But the coordinate  (0, –2) is the lowest point on the graph.

Therefore, (0, –2) is the turning point for the continuous function given.

the rocket cannot penetrate the ground, in a contextual sense, therefore the roots of this equation should be the domain of this function. i'll let you find the roots of this quadratic equation (hint: use the quadratic formula:  

1- c

2-a

step-by-step explanation:

mark brainliest!

the equation is not included but the answer is (1.50, 2.00, 2.75)

step-by-step explanation: