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# Hi good people!,

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Hi good people!,

a real stumper for me?...I have a parabola with equation $$y=-2{x^2}+8x+10$$, and a hyperbola with equation $$y={4 \over x}$$, what i have to find is the point of inetrsection in the 4th quadrant...

I first devided by -2, to simplify the one equation, and then get $$x^2-4x-5$$

Then put the two equations equal to one another, so I get $${4 \over x} =x^2-4x-5$$

Then this goes to $$x^3-4{x^2}-5x=4$$....from here I do not know...please help..

Thank you very much!

Feb 20, 2019

#1
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They don't intersect in the 4th quadrant.

4/x doesn't even have any values in the 4th quadrant.

Feb 20, 2019
edited by Rom  Feb 20, 2019
#2
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i have to find is the point of inetrsection in the 4th quadrant...

There is no intersection in the 4th quadrant, while $$y=\dfrac{4}{x}$$ is only in the 1st and 3th quadrant!

Feb 20, 2019
#3
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Rom and Heureka,

absolutely right, I honestly do apologise for writing this wrong...I was relying on my memory and failed. please forgive me. the quadrant was the 3rd quadrant, and the value of the hyperbola was $$f(g)={16 \over x}$$

would you please show me how to go about the calculations?..so we have:

$$x^3-4{x^2}-5x=16$$

How do I go from here?

juriemagic  Feb 20, 2019
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Hi jm,

$$\begin{array}{|rcll|} \hline y &=& -2x^2+8x+10 \quad & \quad \cdot x \\ yx &=& -2x^3+8x^2+10x \quad & \quad xy = 16 \\ 16 &=& -2x^3+8x^2+10x \quad & \quad :2 \\ 8 &=& -x^3+4x^2+5x \quad & \quad \cdot (-1) \\ -8 &=& x^3-4x^2-5x \\ \mathbf{ x^3-4x^2-5x} &\mathbf{=}& \mathbf{-8} \\ \hline \end{array}$$

I get  $$x^3-4x^2-5x= -8$$, why do you get  $$x^3-4{x^2}-5x=16$$ ?

Feb 20, 2019
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Hello Heureka,

i simplified the equation of the parabola before I set the two equations equal to one another.

juriemagic  Feb 20, 2019
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Hello Heureka,

i simplified the equation of the parabola before I set the two equations equal to one another...oh golly...I see now...that was wrong..because the one is equal to the other, and if you change one of them, then obviously they will not be equal to one another anymore....I see that...

okay, so then it is $$x^3-4{x^2}-5x=-8$$

it really is from here that I just do not know how to get the point of intersection in the 3rd quadrant...can you please help me here?...

juriemagic  Feb 20, 2019
#7
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We have a parabola with equation

$$y=-2{x^2}+8x+10$$,

and a hyperbola with equation

$$xy=16$$

We have to solve  $$\mathbf{ x^3-4x^2-5x +8=0}$$

1.

We attempt x = 1

$$\begin{array}{|rcll|} \hline \mathbf{ x^3-4x^2-5x+8} &\mathbf{=}& \mathbf{0} \quad | \quad x=1 \\\\ 1^3-4\cdot 1^2-5\cdot 1 +8 &\overset{?}{=} & 0 \\ 1 -4 -5 +8 &\overset{?}{=} & 0 \\ 9-9 &\overset{?}{=} & 0 \\ 0 & = & 0\ \checkmark \\ \hline \end{array}$$

We have found the first root, it is $$x = 1 \quad (1.\text{ quadrant })$$

$$\begin{array}{|rcll|} \hline x^3-4x^2-5x+8 = ( ~?~ )(x-1) \\ (x^3-4x^2-5x+8) : (x-1) = ( ~?~ ) \\ \hline \end{array}$$

2.

Polynomial long division:

$$\begin{array}{|rcll|} \hline x^3-4x^2-5x+8 = ( x^2 - 3x-8 )(x-1) &=& 0 \\\\ x^2 - 3x -8 &=& 0 \\ x &=& \dfrac{3 \pm \sqrt{3^2-4(-8) } } {2} \\ x &=& \dfrac{3 \pm \sqrt{9+32} } {2} \\ x &=& \dfrac{3 \pm \sqrt{41} } {2} \\\\ x_2 &=& \dfrac{3 + \sqrt{41} } {2} \\ x_2 &=& 4.70156211872 \quad | \quad 1.\text{ quadrant } \\\\ x_3 &=& \dfrac{3 - \sqrt{41} } {2} \\ \mathbf{ x_3 } &\mathbf{=}& \mathbf{-1.70156211872} \quad | \quad 3.\text{ quadrant } \\\\ y_3 &=& \dfrac{16}{x_3} \\ y_3 &=& \dfrac{16}{-1.70156211872} \\ \mathbf{ y_3 } &\mathbf{=}& \mathbf{-9.40312423743} \\ \hline \end{array}$$

The point of intersection in the 3th quadrant is $$\mathbf{(-1.70156211872, -9.40312423743)}$$

heureka  Feb 20, 2019
edited by heureka  Feb 20, 2019
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Heureka,

got it, got it, got it.......thank you kindly...I do appreciate!

juriemagic  Feb 20, 2019