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Alexxandr [17]
3 years ago
7

Please help I don’t understand this.

Mathematics
1 answer:
tresset_1 [31]3 years ago
3 0

The average rate of change of <em>g(x)</em> over the interval [2, 8] is given by

(<em>g</em> (8) - <em>g</em> (2)) / (8 - 2)

In other words, it's the slope of the line through the points (2, <em>g</em> (2)) and (8, <em>g</em> (8)).

Use the definition of the function to evaluate it at the points in the numerator:

• 8 ≥ 4, so using the second piece, <em>g</em> (8) = -0.5(8) + 8 = 4

• 2 < 4, so <em>g</em> (2) = 5(2) + 1 = 11

Then the average rate of change is

(<em>g</em> (8) - <em>g</em> (2)) / (8 - 2) = (4 - 11) / 6 = -7/6

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Step-by-step explanation:

Vertical asymptote can be Identites if there is a factor only in the denominator. This means that the function will be infinitely discounted at that point.

For example,

\frac{1}{x - 5}

Set the expression in the denominator equal to 0, because you can't divide by 0.

x - 5 = 0

x = 5

So the vertical asymptote is x=5.

Disclaimer if you see something like this

\frac{(x - 5)(x + 3)}{(x - 5)}

x=5 won't be a vertical asymptote, it will be a hole because it in the numerator and denominator.

Horizontal:

If we have a function like this

\frac{1}{x}

We can determine what happens to the y values as x gets bigger, as x gets bigger, we will get smaller answers for y values. The y values will get closer to 0 but never reach it.

Remember a constant can be represent by

a \times  {x}^{0}

For example,

1 = 1 \times  {x}^{0}

2 =  2 \times {x}^{0}

And so on,

and

x =  {x}^{1}

So our equation is basically

\frac{1 \times  {x}^{0} }{ {x}^{1} }

Look at the degrees, since the numerator has a smaller degree than the denominator, the denominator will grow larger than the numerator as x gets larger, so since the larger number is the denominator, our y values will approach 0.

So anytime, the degree of the numerator < denominator, the horizontal asymptote is x=0.

Consider the function

\frac{3 {x}^{2} }{ {x}^{2}  + 1}

As x get larger, the only thing that will matter will be the leading coefficient of the leading degree term. So as x approach infinity and negative infinity, the horizontal asymptote will the numerator of the leading coefficient/ the leading coefficient of the denominator

So in this case,

x =  \frac{3}{1}

Finally, if the numerator has a greater degree than denominator, the value of horizontal asymptote will be larger and larger such there would be no horizontal asymptote instead of a oblique asymptote.

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