(NOT SOLVED YET) Advanced Function question (need help)

[peeky47]

Question-1 hosted at ImgBB

Image Question-1 hosted in ImgBB

ibb.co

Picture of question^

Need help as this is very confusing.

How can there be a Horizontal asymptote at y=2 as it look like there is an x-value at y=2

And subbing (0,0) gives a value of 0 for c, but c cannot be zero as stated in the question and thus my confusion.

Your help would be appreciated.

Thanks

 

[MarkFL]

Here's the image:

 

[Steven G]

Question-1 hosted at ImgBB

Image Question-1 hosted in ImgBB

ibb.co

Picture of question^

Need help as this is very confusing.

How can there be a Horizontal asymptote at y=2 as it look like there is an x-value at y=2

And subbing (0,0) gives a value of 0 for c, but c cannot be zero as stated in the question and thus my confusion.

Your help would be appreciated.

Thanks

Please show your work showing that (0,0) is a point on the curve implies that c=0, as that is not correct.

How can there be a Horizontal asymptote at y=2 as it look like there is an x-value at y=2? Why do you think that a curve can't can't cross a horizontal asymptote. I think that you think the same should be true for horizontal asymptotes as with vertical asymptotes. You are correct that a function graph should not cross a vertical asymptote but it can cross a horizontal asap. You should look at the function y = (sin(x))/x. All that a horizontal asymptote says is that as x approaches infinity or as x approaches negative infinity the function will get closer and closer to a particular y value.

Click on the link here to see the graph y= sin(x)/x. Note how x gets larger the y-vales are approaching 0 and crossing the line y= 0!

 

[Li Batton]

I tried this for part ii.

From part i we know that a = 4. From the graph we know that there is
a value of x for which the function has a y value of 2.
Therefore, we can write

[MATH]2=\frac{4x^2+bx}{2x^2-c}[/MATH]
Now solving for c

[MATH]2(2x^2-c)=4x^2+bx[/MATH]
[MATH]4x^2-2c=4x^2+bx[/MATH]
[MATH]-2c=bx[/MATH]
[MATH]c=\frac{bx}{-2}[/MATH]
c is a negative number therefore you have shown that c < 0

 

[Steven G]

I tried this for part ii.

From part i we know that a = 4. From the graph we know that there is
a value of x for which the function has a y value of 2.
Therefore, we can write

[MATH]2=\frac{4x^2+bx}{2x^2-c}[/MATH]
Now solving for c

[MATH]2(2x^2-c)=4x^2+bx[/MATH]
[MATH]4x^2-2c=4x^2+bx[/MATH]
[MATH]-2c=bx[/MATH]
[MATH]c=\frac{bx}{-2}[/MATH]
c is a negative number therefore you have shown that c < 0

Not correct-sorry. Why can't b<0 and c>0? There is no reason at all to assume that b>0! Were you at least assuming that x>0 from looking at the graph?

There is a simple reason why c<0. Go ahead and try again.

 

[Li Batton]

Yes, I assumed b was positive. Which, as you point out, is not correct.

 

[MarkFL]

Suppose you write the given function as:

[MATH]f(x)=\frac{a+\dfrac{b}{x}}{2-\dfrac{c}{x^2}}[/MATH] where \(x\ne0\)

Now as \(x\to\pm\infty\) what value must \(f(x)\) approach?

We see by the absence of vertical asymptotes that the denominator has no real roots, which means its discriminant must be negative. What does this imply?

 

[Li Batton]

Suppose you write the given function as:

[MATH]f(x)=\frac{a+\dfrac{b}{x}}{2-\dfrac{c}{x^2}}[/MATH] where \(x\ne0\)

Now as \(x\to\pm\infty\) what value must \(f(x)\) approach?

We see by the absence of vertical asymptotes that the denominator has no real roots, which means its discriminant must be negative. What does this imply?

Oh!! I see it now. Thank you