Horizontal Asymptotes, need help test tommorow

[thelazyman]

I understand how they got the horizontal and vertical asymptotes, but I do not know how they got the y and x intercept.

The function is:

y = F(x)= 3 + 2/ root x - 2

I do not understand how they got 16/9 as an x value when y is = to zero., someone please help.

 

[skeeter]

you really need to use grouping symbols to make your expression clear for others who read it.

do you mean ...

\(\displaystyle \L f(x) = 3 + \frac{2}{\sqrt{x-2}}\)

or

\(\displaystyle \L f(x) = 3 + \frac{2}{\sqrt{x} - 2}\)

or something else entirely :?:

in any case ...
to find the y-intercept, calculate f(0).
to find the x-intercept, set f(x) = 0 and solve for x.

 

[thelazyman]

Yeah that is the question. and it was when they make y = 0, what happens to the x value. dunno how they got 16/9 to make y = 0

 

[stapel]

Yeah that is the question.

Which one? The tutor posted two different options.

it was when they make y = 0, what happens to the x value.

Are you referring to finding the x-intercepts of f(x)...?

dunno how they got 16/9 to make y = 0

Did you try plugging 16/9 in for x and seeing what you got?

Eliz.

 

[galactus]

They used algebra.

\(\displaystyle \L\\3+\frac{2}{\sqrt{x}-2}=0\)

Multiply by the LCD, which is \(\displaystyle \sqrt{x}-2\)

\(\displaystyle \L\\3(\sqrt{x}-2)+\sout{(\sqrt{x}-2)}\cdot\frac{2}{\sout{\sqrt{x}-2}}=0\)

\(\displaystyle \L\\3(\sqrt{x}-2)+2=0\)

\(\displaystyle \L\\3(\sqrt{x}-2)=-2\)

\(\displaystyle \L\\\sqrt{x}-2=\frac{-2}{3}\)

\(\displaystyle \L\\\sqrt{x}=\frac{-2}{3}+2\)

\(\displaystyle \L\\x=(\frac{4}{3})^{2}=\frac{16}{9}\)