Need help solving this problem

[EddyBenzen122]

[math]let\ f(x)\ =\ \frac{8x-5}{-2x+6},\ x\neq3[/math]
The line y = k, where [imath]k\in R[/imath] intersects the graph of [imath]\left|f\left(x\right)\right|[/imath] at exactly one point. Find the possible values of k.

My attempt:
I was thinking discriminant so this is what my work looks like.

[imath]k=\left|\frac{8x-5}{-2x+6}\right|[/imath], [imath]\pm k=\frac{8x-5}{-2x+6}[/imath]
[imath]8x-5=-2kx+6k[/imath], [imath]8x-5=2kx\ -\ 6k[/imath]
[imath]\mathrm{\Delta}_1=\left(2k+8\right)^2-4\left(0\right)\left(-5-6k\right)=4k^2\ \ 32k+64\ =\ 4\left(k^2\ +\ 8k+16\right)[/imath]
[imath]\mathrm{\Delta}_2=\left(-2k+8\right)^2-4\left(0\right)\left(-5+6k\right)=4k^2\ -32k+64\ =\ 4\left(k^2-\ 8k+16\right)[/imath]
[imath]0\ =\ 4\left(k^2\ +\ 8k+16\right)\ =4\ \left(k+4\right)^2[/imath]
[imath]0\ =\ 4\left(k^2\ -\ 8k+16\right)\ =\ \left(k\ -\ 4\right)^2[/imath]
[imath]k=\pm4[/imath]
The answer key says k = 0 and k = 4, so I have no idea what I've done wrong.

 

[Harry_the_cat]

Can you explain why you are using the discriminant of a quadratic? The equation is not a quadratic.

I would consider what the graph of y=f(x) and y=abs(f(x)) looks like.

 

[BigBeachBanana]

[math]let\ f(x)\ =\ \frac{8x-5}{-2x+6},\ x\neq3[/math]
The line y = k, where [imath]k\in R[/imath] intersects the graph of [imath]\left|f\left(x\right)\right|[/imath] at exactly one point. Find the possible values of k.

My attempt:
I was thinking discriminant so this is what my work looks like.

[imath]k=\left|\frac{8x-5}{-2x+6}\right|[/imath], [imath]\pm k=\frac{8x-5}{-2x+6}[/imath]
[imath]8x-5=-2kx+6k[/imath], [imath]8x-5=2kx\ -\ 6k[/imath]
[imath]\mathrm{\Delta}_1=\left(2k+8\right)^2-4\left(0\right)\left(-5-6k\right)=4k^2\ \ 32k+64\ =\ 4\left(k^2\ +\ 8k+16\right)[/imath]
[imath]\mathrm{\Delta}_2=\left(-2k+8\right)^2-4\left(0\right)\left(-5+6k\right)=4k^2\ -32k+64\ =\ 4\left(k^2-\ 8k+16\right)[/imath]
[imath]0\ =\ 4\left(k^2\ +\ 8k+16\right)\ =4\ \left(k+4\right)^2[/imath]
[imath]0\ =\ 4\left(k^2\ -\ 8k+16\right)\ =\ \left(k\ -\ 4\right)^2[/imath]
[imath]k=\pm4[/imath]
The answer key says k = 0 and k = 4, so I have no idea what I've done wrong.

Since [imath]|f(x)|\ge 0[/imath], so you know that [imath]k=-4[/imath] is wrong.

 

[EddyBenzen122]

Since [imath]|f(x)|\ge 0[/imath], so you know that [imath]k=-4[/imath] is wrong.

ok but why does k = 0 as well?

 

[EddyBenzen122]

Can you explain why you are using the discriminant of a quadratic? The equation is not a quadratic.

I just thought discriminant would allow us to find possible values of a variable when two graphs intersect each other. If you believe discriminant doesn't work, could you provide another way of solving this problem?

 

[BigBeachBanana]

ok but why does k = 0 as well?

In general, [imath]|f(x)|[/imath] has a vertex where [imath]-f(x)[/imath] gets"flipped" into positive [imath]f(x)[/imath] due to the absolute value. At this point the function will be one-to-one. In this case, it lies at the point [imath]\left(\frac{5}{8},0\right) \implies y=0[/imath]. I'll let you show it as an exercise.

 

[Harry_the_cat]

I just thought discriminant would allow us to find possible values of a variable when two graphs intersect each other. If you believe discriminant doesn't work, could you provide another way of solving this problem?

draw the graph

 

[Steven G]

I 2nd Draw The Graph.
I drew the graph and the answer was obvious.