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Find the points of intersection of the two parabolas with equation y=-(x-3)^2+2  and y=x^2-4 x+1  .

Option: 1

(3, \sqrt{2})$ and $(\sqrt{2}, 2)


Option: 2

(0,3+\sqrt{2})$ and $(2,3+\sqrt{2})


Option: 3

(0,2)$ and $(2,1)


Option: 4

(3+\sqrt{2}, 0)$ and $(3-\sqrt{2}, 2)


Answers (1)

best_answer

To find the points of intersection of the two parabolas, we need to set their equations equal to each other and solve for x:

\Rightarrow-(x-3)^2+2=x^2-4 x+1

After simplifying and expanding the square, we obtain:

-x^2+6 x-7=0

Using the quadratic formula, we can solve this quadratic equation:

\Rightarrow x=\frac{\left(-b \pm \sqrt{\left(b^2-4 a c\right)}\right)}{(2 a)}

This time, the values are a=-1,b=6 \: and\: c=-7. These values are entered into the quadratic formula as follows:

\begin{aligned} & x=\frac{\left(-6 \pm \sqrt{\left(6^2-4(-1)(-7)\right)}\right)}{(2(-1))} \\ & x=\frac{(-6 \pm \sqrt{(36-28)})}{(-2)} \\ & x=\frac{(-6 \pm \sqrt{8})}{(-2)} \end{aligned}

If we simplify, we get:

x=3 \pm \sqrt{2}

So, the x-coordinates of the points of intersection are x=3+\sqrt{2} \text { and } x=3-\sqrt{2} \text {. }

We can enter these x values into either of the original equations to determine the ycoordinates. Let's employ the first equation:

\Rightarrow y=-(x-3)^2+2......(1)\\\\ Inputting\: \: x=3+\sqrt{2},

\begin{aligned} & y=-((3+\sqrt{2})-3)^2+2 \\ & y=-\sqrt{2}^2+2 \\ & y=0 \end{aligned}

Therefore, (3+\sqrt{2}, 0) is one point of intersection.

Inputting  x=3-\sqrt{2} in equation (1)

\begin{aligned} & y=-((3-\sqrt{2})-3)^2+2 \\ & y=-(-\sqrt{2})^2+2 \\ & y=2 \end{aligned}

The opposite intersection is therefore (3-\sqrt{2}, 2)

The two parabolas therefore meet at the coordinates (3+\sqrt{2}, 0) \text { and }(3-\sqrt{2}, 2) \text {. } 

 

 

 

Posted by

seema garhwal

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