PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!!
1 answer:
Answer: (A) vertical asymptote: x = 2, horizontal asymptote: y = 1
<u>Step-by-step explanation:</u>
<u>Vertical Asymptote</u> is the restriction on the x-value. The denominator cannot be zero, so x - 2 ≠ 0 ⇒ x ≠ 2
The restricted value on x is when x = 2 <em>which is the vertical asymptote</em>
<u>Horizontal Asymptote</u> (H.A.) is the restriction on the y-value. This is a comparison of the numerator (n) and denominator (m). There are 3 rules that will help you:
- n > m No H.A. (use long division to find slant asymptote)
- n = m H.A. is the coefficient of n divided by coefficient of m
- n < m H.A. is 0
In the given problem, n < m so y = 0, however there is also a vertical shift of up 1 so the H.A. also shifts up. This results in H.A. of y = 1
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3 to the negative second power is written like this: 
.
When simplifying an expression with a negative exponent, you take a positive version of the exponent, in this case that would be 2, and apply that to the base.
After doing that, we have 9.
The next step is to put one over that number.
In this case, now after doing that the answer is
.
Hope this helps!
When simplifying an expression with a negative exponent, you take a positive version of the exponent, in this case that would be 2, and apply that to the base.
After doing that, we have 9.
The next step is to put one over that number.
In this case, now after doing that the answer is
Hope this helps!
Answer: (a) e ^ -3x (b)e^-3x
Step-by-step explanation:
I suggest the equation is:
d/dx[integral (e^-3t) dt
First we integrate e^-3tdt
Integral(e ^ -3t dt) as shown in attachment and then we differentiate the result as shown in the attachment.
(b) to differentiate the integral let x = t, and substitute into the expression.
Therefore dx = dt
Hence, d/dx[integral (e ^-3x dx)] = e^-3x
