Hollymae764
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Can someone help me out with part d please

I set up the equation like this:

-5/3 (x+4) + 8 = 5/3 x + k
and got -10/3 x + 4/3 - k

then i did b^2 - 4ac
to get k < 4/3

then i did the same but with (-x-4) and i couldn't solve since the x's just cancelled out.


I know this is completely wrong so can someone please help
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mqb2766
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The equation is not quadratic, so you can't use the discriminant.
You could plot the line (and function) for different values of k and think about why the two gradients 5/3 are the same, as it makes it easy to determine the critical value of k.
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Hollymae764
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(Original post by mqb2766)
The equation is not quadratic, so you can't use the discriminant.
You could plot the line (and function) for different values of k and think about why the two gradients 5/3 are the same, as it makes it easy to determine the critical value of k.
Okay so I get that the graph must be to the left of -8.8 which is the point the f(x) intersects with the x axis. Ive used desmos to find a rough point for K but how do i do this algebraically?
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mqb2766
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(Original post by Hollymae764)
Okay so I get that the graph must be to the left of -8.8 which is the point the f(x) intersects with the x axis. Ive used desmos to find a rough point for K but how do i do this algebraically?
k is a vertical shift of the line (the y intercept).
How will this relate to the "left line" of f(x) when x < -4?
Link your desmos graph if unsure.
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Hollymae764
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(Original post by mqb2766)
k is a vertical shift of the line (the y intercept).
How will this relate to the "left line" of f(x) when x < -4?
Link your desmos graph if unsure.
im a bit confused as to what you mean
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Hollymae764
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(Original post by Hollymae764)
im a bit confused as to what you mean
so i get that it is a vertical shift, but what does x <-4 have to do with it?
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mqb2766
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(Original post by Hollymae764)
im a bit confused as to what you mean
It should be obvious from the desmos graph? Can you share/link your graph?
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mqb2766
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(Original post by Hollymae764)
so i get that it is a vertical shift, but what does x <-4 have to do with it?
Your function f(x) has two lines. One is defined for x<-4?
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Hollymae764
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(Original post by mqb2766)
It should be obvious from the desmos graph? Can you share/link your graph?
https://www.desmos.com/calculator/vsw7eqawuf
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Hollymae764
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(Original post by mqb2766)
Your function f(x) has two lines. One is defined for x<-4?
oh i see what youre saying the left line is defined by x<-4 but how does this help me with the answer. Sorry for being slow this is just a bit confusing to me
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mqb2766
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(Original post by Hollymae764)
oh i see what youre saying the left line is defined by x<-4 but how does this help me with the answer. Sorry for being slow this is just a bit confusing to me
Just shift the blue line (change k) up and down a bit. When are there no solutions (intersect with the red curve) and how does this relate to the increasing part (red line) of f(x).?
In your slider, make the max value of k about 20.
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Hollymae764
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i got the answer k > 44/3

i did this by looking at the root of f(x) defined by x < -4

I did 5/3 (8.8) + k = 0
k = 44/3 and since the k is a vertical shift I know if i increase k then line line will move upwards and away from f(x) so
k > 44/3

is that correct?
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mqb2766
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(Original post by Hollymae764)
i got the answer k > 44/3

i did this by looking at the root of f(x) defined by x < -4

I did 5/3 (8.8) + k = 0
k = 44/3 and since the k is a vertical shift I know if i increase k then line line will move upwards and away from f(x) so
k > 44/3

is that correct?
You can test this on your graph, but yes. 44/3 is the y intercept of the red line of interest. The blue line must be above this for no intersection points.
Your explanation could be a bit clearer.
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