Question

PLEASE HELP URGENT EXPLAIN HOW TO DETERMINE IF ITS INCREASING OR DECREASING

Answer 1

Answer:

Decreasing from 0 to infinity.

Step-by-step explanation:

Write the function:

$h(x)=-5\sqrt{x}$

If you look and analyze the function, we already can tell that it doesn't admit negative numbers, because you can't take the square root of a negative number. Theferore, all the options that go from -∞ to 0 are not correct, because the function doesn't even generate values in that interval.

Then, to see of the function increases or decreases from 0 to ∞, just evaluate the function, at least 3 times, to see where the values of y go.

Let's take 3 arbitrary values for this:

$h(0)=-5\sqrt{0} =0\\\\h(0)=-5\sqrt{49} =-35\\\\h(0)=-5\sqrt{15000} =-612.372$

With these values, we can clearly tell that the function is decreasing from 0 to infinity.

A more analytic way to determine this would be by finding the average rate of change of the function on an interval that goes from 0 to any number but infinity.

Average rate of change formula: $A(x)=\frac{h(b)-h(a)}{b-a}$

Take arbitrary values and substitute the function:

$A(x)=\frac{h(50)-h(5)}{50-5}\\\\A(x)=\frac{-35.35-(-11.18)}{45}\\\\A(x)=\frac{-35.35+11.18}{45}\\\\A(x)=\frac{-24.17}{45}\\\\A(x)= -0.5364$

The result is -0.5364, this means that the average change or the function from x=5 to x=50 is negative, and this indicates that the function is decreasing.

Answer 2

Answer:

The function is decreasing on the interval (0, ∞).

Step-by-step explanation:

Definitions

• Domain: Set of all possible input values (x-values).
• A function is said to be increasing if the y-values increase as the x-values increase.
• A function is said to be decreasing if the y-values decrease as the x-values increase.

Given function:

$h(x)=-5\sqrt{x}$

The parent function of the given function is:

$y = \sqrt{x}$

As we cannot square root a negative number, the parent function has a restricted domain of [0, ∞).  

As  $\sqrt{x}\geq 0$  the parent function is increasing on the interval (0, ∞).

Transformations

$y=a\:f(x) \implies f(x) \: \textsf{stretched parallel to the y-axis (vertically) by a factor of}\:a$

$y=-f(x) \implies f(x) \: \textsf{reflected in the} \: x \textsf{-axis}$

Therefore, the parent function has been stretched vertically by a factor of 5 and reflected in the x-axis to become function h(x).

Therefore, the domain remains the same.  

The stretch does not affect the increasing/decreasing nature of the function, but as the function is reflected in the x-axis, it is now decreasing on the interval (0, ∞).