Question

HELP ME ENTERNAME(In-Game Friend)!! Let's say line $l$ that $x-2y+a=0$ is transferred by parallel movement from $y=x^2-2x$ to $y=x^2-10x+20$. If line $l$ passes the origin, find the value of the constant a.

Answer 1

Answer:

12

Step-by-step explanation:

Hi :)

The first parabola is $y=x^2-2x$ and it simplifies into $y=(x-1)^2-1$.

The moved graph is $y=x^2-10x+20$ and it simplifies into $y=(x-5)^2-5$.

To see the parallel movement, let's slightly change the form.

$y=(x-1)^2-1$ → $y+1=(x-1)^2$

$y=(x-5)^2-5$ → $y+5=(x-5)^2$

We can observe it is f(x, y) → f(x-4, y+4).

So the line $l$ becomes $(x-4)-2(y+4)+a=0$. It simplifies into the linear graph of $x-2y+a-12=0$.

It is required that (0, 0) is on the line so that $l$ passes the origin.

Now, we put in the value.

$a-12=0$ and so we get a value 12.

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Answer 2

Answer:

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