Does f(x) = kx^2 - (6k-5)x + 8k + 7 pass through (a,b) and (c,d)?

[gerimis]

Homework Statement

Graphic Function f(x) = kx^2 - (6k-5)x + 8k + 7 always pass through (a,b) and (c,d). So, b + d = …

 

[JonF]

if f(x) passes through (a,b) then you know:
b=k*a² - (6k-5)a + 8k + 7

do you see how this helps?

 

[gerimis]

if f(x) passes through (a,b) then you know:
b=k*a² - (6k-5)a + 8k + 7

do you see how this helps?

Ya, but I am still not understand, because the answer is Real Number.

 

[JonF]

do what i did to b with d and add them both together, you will end up with a real number in terms of a,c, and k

 

[gerimis]

I got:
b + d = k(a² + c²) - {(6k - 5)*(a + c)} + 16k + 14

And i don't know what I've to do with that. :D

 

[JonF]

that is a solution

 

[gerimis]

I don't understand :(
Because the answer options:
A. 19
B. 24
C. 29
D. 34
E. 39

 

[Apphysicist]

Do you know anything about k that would restrict it beyond "an element of the reals?"

 

[Mark44]

My sense is that there is some information missing. gerimis, are you sure you've provided all the information here?

 

[gerimis]

Do you know anything about k that would restrict it beyond "an element of the reals?"

No, i don't...
I have tried to use Discriminant D>0, but i don't still get the 'k'.

 

[Mentallic]

You can't possibly get an answer that doesn't depend on some of the variables. This is because, say we gave even more information such as the points are (-1,b) and (1,d) so we don't have a and b anymore and the quadratic is y=kx² then we will still have the value of b+d being dependent on the value of k in the quadratic. We have b+d=2k so if k=1, then the quadratic y=x² passes through (-1,1) and (1,1) so b+d=1+1=2. If k=2, b+d=4 etc.

 

[Citan Uzuki]

The problem is saying that for any value of k, the function defined by f(x) = kx^2 - (6k-5)x + 8k + 7 passes through the points (a, b) and (c, d). This means that f(a) and f(c) do not depend on the value of k.

Now, rewriting f(x) as k(x^2 - 6x + 8) - 5x + 7, what must be true of a and c, for f(a) and f(c) not to depend on the value of k?

 

[Mentallic]

This means that f(a) and f(c) do not depend on the value of k.

Yes they do...

 

[Citan Uzuki]

Yes they do...

There are two specific numbers a and c for which they do not. Look at the equation f(x) = k(x^2 - 6x + 8) - 5x + 7 again.

 

[Mentallic]

There are two specific numbers a and c for which they do not. Look at the equation f(x) = k(x^2 - 6x + 8) - 5x + 7 again.

Oh yes I just skimmed over your second paragraph. That specific quote however is incorrect because you cannot make that false conclusion. As you said, there are specific values of a and c such that b+d is independent of k. Maybe this is what was required of the OP?

 

[HallsofIvy]

I agree that the phrasing "for any value of k", implies that the values of a, b, c, and d must not depend on k. If that is true then we must have,
taking k= 0, $b+ d= 5a- 5c+ 14$
taking k= 1, $b+ d= a^2+ c^2- a- c+ 30$
taking k= -1, $b+ d= -a^2- c^2+ 21a+ 21c- 2$
taking k= 2, $b+ d= 2a^2+ 2c^2- 27a- 27c+ 46$

That gives 4 equations to solve for a, b, c, and d. Since the question only asks for b+ d you may not have to solve completely.

For example, adding the second and third equation gives $2(b+ d)= 19a+ 19c+ 28$ while adding twice the third equation to the fourth gives 2(a+ b)= 14a- 14c+ 42. Those, together with b+ d= 5a- 5c+ 14 may allow you to solve for b+ d directly.