# Function question

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#1
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 0
4 weeks ago
#2
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.
Last edited by mqb2766; 4 weeks ago
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#3
(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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4 weeks ago
#4
(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?
Last edited by mqb2766; 4 weeks ago
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#5
(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?
im a bit confused as to what you mean
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#6
(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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4 weeks ago
#7
(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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4 weeks ago
#8
(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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#9
(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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#10
(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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4 weeks ago
#11
(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).?
Last edited by mqb2766; 4 weeks ago
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#12
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?
0
4 weeks ago
#13
(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.
Last edited by mqb2766; 4 weeks ago
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