# Thread: A line with equation x = p

1. ## A line with equation x = p

Given is the function fp(x) = 4/x + p With p element of R

The line with equation x = p intersects the graph of fp(x) in point Sp. Determine the values of the y-coordinate of Sp can take on.

I come to the next, but that is not good, according to the answer sheet , that gives: Bsp = (<=,-4] and [4,=>)

I substitute for x p => 4/p + p =0 => 4 + p^2=0 => p= +-4

Who can see what goes wrong with my development?

2. ## Re: A line with equation x = p

What you have written is y= (4/x)+ p. Sometimes that is, incorrectly, written when a person really means y= 4/(x+ p). Can you verify which you mean?

Why are you setting y equal to 0? The problem asks for possible values of y, not p.

If y= (4/x)+ p and x= p then y= (4/p)+ p= (4+p^2)/p. If we attempt to solve that for p, we get 4+ p^2= py or p^2- py+ 4= 0. The "discriminant" from the quadratic formula is p^2- 16. p will be real if and only if $p^2- 16\ge 0$ or $p^2\ge 16$. That gives $p\le -4$ or $p\ge 4$.

3. ## Re: A line with equation x = p

Originally Posted by HallsofIvy
What you have written is y= (4/x)+ p. Sometimes that is, incorrectly, written when a person really means y= 4/(x+ p). Can you verify which you mean?

Why are you setting y equal to 0? The problem asks for possible values of y, not p.

If y= (4/x)+ p and x= p then y= (4/p)+ p= (4+p^2)/p. If we attempt to solve that for p, we get 4+ p^2= py or p^2- py+ 4= 0. The "discriminant" from the quadratic formula is p^2- 16. p will be real if and only if $p^2- 16\ge 0$ or $p^2\ge 16$. That gives $p\le -4$ or $p\ge 4$.

As you wrote the function is correct. I should have written (4/x)

I do not know how you come at p^2- 16. Can you further explain in detail? Or should I just know it?

4. ## Re: A line with equation x = p

The "quadratic formula", $x= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$ gives the roots of the quadratic equation $ax^2+ bx+ c= 0$. x will be real as long as the "discriminant", $b^ 2- 4ac$, is not negative. That is, $b^2- 4ac\ge 0$.

Here, the equation for p is the quadratic equation $p^2- yp+ 4= 0$ so that, in the formula above, a= 1, b= -y and c= 4. The discriminant, that determines whether there are any real roots is $(-y)^2- 4(1)(4)= y^2- 16$. The condition that there be real roots is that $y^2- 16= (x- 4)(x+ 4)$ be non-negative. In order that the product of two numbers be non-negative, they either be both non-negative themselves or both negative.

5. ## Re: A line with equation x = p

Hey HallsofIvy

thanks for your explanation! I get it now.