# Function question

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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

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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#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.

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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(Original post by

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.

**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.

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#4

(Original post by

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?

**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?

How will this relate to the "left line" of f(x) when x < -4?

Link your desmos graph if unsure.

Last edited by mqb2766; 4 weeks ago

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(Original post by

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.

**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.

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(Original post by

im a bit confused as to what you mean

**Hollymae764**)im a bit confused as to what you mean

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#7

(Original post by

im a bit confused as to what you mean

**Hollymae764**)im a bit confused as to what you mean

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#8

(Original post by

so i get that it is a vertical shift, but what does x <-4 have to do with it?

**Hollymae764**)so i get that it is a vertical shift, but what does x <-4 have to do with it?

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(Original post by

It should be obvious from the desmos graph? Can you share/link your graph?

**mqb2766**)It should be obvious from the desmos graph? Can you share/link your graph?

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(Original post by

Your function f(x) has two lines. One is defined for x<-4?

**mqb2766**)Your function f(x) has two lines. One is defined for x<-4?

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#11

(Original post by

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

**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

In your slider, make the max value of k about 20.

Last edited by mqb2766; 4 weeks ago

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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?

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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#13

(Original post by

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?

**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?

Your explanation could be a bit clearer.

Last edited by mqb2766; 4 weeks ago

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