Function limit problem.

[pinkcalculator]

I don't even know where to start with this. Yuck.
Suppose you have the function f(x)= 2x+b/cx+d. This function is undefined at x= -4/3. The graph of the function strikes the x axis at 8 and strikes the y axis at 1. What are the values for b, c, and d.

I think this is right?
If it's undefined at x=-4/3, using that information to set the denominator to zero
c(-4/3) + d=0
-4/3c+d=0.
Can I do that? Or am I supposed to set the ENTIRE fraction with that x value?
Like this
2(-3/4) +b/-4/3c+d=0?

Any suggestions would be greatly appreciated.

 

[Deleted member 4993]

I don't even know where to start with this. Yuck.
Suppose you have the function f(x)= 2x+b/cx+d. This function is undefined at x= -4/3. The graph of the function strikes the x axis at 8 and strikes the y axis at 1. What are the values for b, c, and d.

I think this is right?
If it's undefined at x=-4/3, using that information to set the denominator to zero
c(-4/3) + d=0
-4/3c+d=0.
Can I do that? Or am I supposed to set the ENTIRE fraction with that x value?
Like this
2(-3/4) +b/-4/3c+d=0?

Any suggestions would be greatly appreciated.

Again you need to use grouping symbols (parenthesis) - to properly indicate order of operations. From the rest of the question, I surmise, you have:

\(\displaystyle f(x) = \frac{2x+b}{cx+d}\)

if that is so - then you should have posted:

f(x) = (2x+b)/(cx+d)

then your first step is correct ? d = 4/3 * c

The graph of the function strikes the x axis at 8

that means f(8) = 0 ? b = -16

strikes the y axis at 1

that means f(0) = 1 ? ??

 

[pinkcalculator]

f(0)=1,
-4/3c+=1
so c=-3/4!

Thanks for the help!

 

[Deleted member 4993]

f(0)=1,
-4/3c+=1
so c=-3/4!

Thanks for the help!

f(x) = (2x+b)/(cx+d)

f(0) = 1

1 = (2*0+b)/(c*0+d)

1 = b/d

d = b = -16

c = 3/4 * d = 3/4*(-16) = -12