3rd degree function has a line y = px + q that tangens at A and intersects at B

[Flible]

Hi there,

I'm a bachelor student of something entirely unrelated to mathematics, but I do like to do it every once in a while. This is an assignment from a Belgian University entrance exam. I believe my mistake is a lack of algebraic insight; it's been 4 years, I have the brains but just not the tools anymore I think

f(x) = x^3 - 11x^2 - 25x - 13
y = p*x + q
Line Y is the tangent of f(x) at A(a,f(a)) and intersects with f(x) at B (13,0). What is the sum of p+q?

I can't help but be prejudiced about the answer being 0, but I can't come to that answer with algebra. Help!

My attempt:
For A, f(a) = y(a) ∧ f ' (a) = y' (a)
a^3 - 11a^2 - 25a - 13 = p * a + q ∧ 3a^2 - 22a - 25 = p

Substitute for p

a^3 - 11a^2 - 25a - 13 = a(3a^2 - 22a - 25) = 3a^3 - 22a^2 - 25a + q
Now q can be expressed in a, and I tried to then substitute both p and q in 0 = p*13 + q, but this didn't work out so well.

 

[skaa]

Let's think that A!=B. So, the equation f(x)-(px+q)=0 must have just two solution, one of which is 13.
q=-13p.
So, let's divide the polynomial x^3-11x^2-25x-13-px+13p[=x^3-11x^2-(25+p)x-13+13p] by x-13.
The quotient is x^2+2x+(1-p). Hence 1-p=1 (to be the quotient a full square). Hence p=q=0.

 

[Flible]

Let's think that A!=B. So, the equation f(x)-(px+q)=0 must have just two solution, one of which is 13.
q=-13p.
So, let's divide the polynomial x^3-11x^2-25x-13-px+13p[=x^3-11x^2-(25+p)x-13+13p] by x-13.
The quotient is x^2+2x+(1-p). Hence 1-p=1 (to be the quotient a full square). Hence p=q=0.

Hi!

Thanks for your reply. I would never have thought of this! There are steps that I can follow but not comprehend, and I like comprehending things fully.
I forgot to add the given information that indeed A != B .

f(x)-(px+q) having 2 solutions sounds to me like one has to assume that the tangent and the intersection are the only two times the two formula's interact, but this doesn't have to be the case (well, obviously it does, but I don't see how one can come to that conclusion?)

f(x) = y --> x^3 - 11x^2 - 25x - 13 - (px + q) = 0
11x^2 - 25x - 13 - px +13p = 0
11x^2 - x(25+p) - 13 + 13p = 0

I've lost you at the point where you try to divide this by (x-13) and come to x^2+2x+(1-p); can you show me the steps?

 

[skaa]

Hi!

This is the division (or factoring - in our case it is the same):
x^3-11x^2-(25+p)x-13+13p=(x^2+2x+1-p)(x-13) (you can multiply, and you will see)

Our polynomial is
f(x)=x^3-11x^2-25x-13
The line:
y(x)=0x+0

Let's check for A(13,0):
polynomial f(13)=0, line y(13)=0 - common point.

Let's check for B(-1,0):
polynomial f(-1)=0, line y(-1)=0 - common point.

Let's check for slope in B(-1,0):
polynomial f'(-1)=0, line slope is 0, too, so the line is a tangent of f(x) in A.

 

[Flible]

Hi!

This is the division (or factoring - in our case it is the same):
x^3-11x^2-(25+p)x-13+13p=(x^2+2x+1-p)(x-13) (you can multiply, and you will see)

Our polynomial is
f(x)=x^3-11x^2-25x-13
The line:
y(x)=0x+0

Let's check for A(13,0):
polynomial f(13)=0, liney(13)=0 - common point.

Let's check for B(-1,0):
polynomial f(-1)=0, line y(-1)=0 - common point.

Let's check for slope in B(-1,0):
polynomial f'(-1)=0, line slope is 0, too, so the line is a tangent of f(x) in A.

Hello!

I'm grateful for your insight! I can follow you up until the point where the quotient is identified: (x^2+2x+1-p). We're getting there!
How did you obtain 1-p = 1? I would think that, since we started off with f(x)=y, and we moved y to f(a)-y=0, we'd be off trying to solve (for a), a^2+2a+1-p = 0?
This can at best mean that 1-p = or < 0, or there are no solutions.

/edit I expanded a bit of my own attempt, now trying to solve:
a^3 - 25a^2 + 143a + 169 = 0
I simply substituted q for -13p and substituted p for f'a; threw all to the left and divided by 2.
I have learned that this can be done with the rational root theorem, something that is not in my country's secondary school curriculum. So I'm still hoping that somebody can explain to me how we can conclude that 1-p = 0 (I assume this is done to have x^2 + 2x + 1-p with only one solution (2^2 - 4*1*(1-p) = 0, where the answer of p=0=q rests. Poq!)