# I know Ihave more than one question but I really needed some serious help on thses last ones. The last three are the hardest and I am lost. 43) The radius of a circle is (x-3) inches. What is...

I know Ihave more than one question but I really needed some serious help on thses last ones. The last three are the hardest and I am lost.

43) The radius of a circle is (x-3) inches. What is the area of the circle? You can express your answer in terms of ∏. Be sure to simplify your answer and include units in your answer.

**44a) **Given the graph of y = -2 + 1/(x-3). What is the x value of the vertical asymptote? Why does y approach infinity for this x value?

**44b) **What is the horizontal asymptote? Why does y approach this value when x gets large?

Yes the last two go together...please help me:) thank you so much!

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Part (3): We can say the line y = c is a horizontal asymptote if

lim -2 + 1/(x-3) = c

(x -> +infinity)

OR

lim -2 + 1/(x-3) = c

(x -> -infinity)

In this example

lim -2 + 1/(x-3) = -2

(x -> +infinity)

And

lim -2 + 1/(x-3) = -2

(x -> -infinity)

Thus the horizonal asymptote is -2.

y approaches this value because when x reaches +inifinity or - inifinity, the term 1/(x-3) reaches zero.

Part (2) x value of the vertical asymptote is 3.

a is said to be a vertical asymptote of the graph of -2 + 1/(x-3) if

limit -2 + 1/(x-3) = +infinity

(x--> a+)

or

limit -2 + 1/(x-3) = - infinity

(x--> a-)

Since the function limit (x--> a+) -2 + 1/(x-3) will reach infinity when the denominator , i.e., (x-3) approches zero, this value is 3.

Thus, vertical asymptote is 3.

For this value y approaches infinity as the value makes a division of 1 by a very small number as x approached 3.

I will only answer one question.

43) A=Pi(x-3)^2

A=Pi(X^2-6x+9)

A=PiX^2-Pi6x+9Pi

.:. This question can not really be done as we need to know the value of x.