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LUCKY_DIMON [66]
3 years ago
8

If I did this wrong pls help me !

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
3 0
Always try to simplify prior graphing. However you have to mention where the function is continued or discontinued.

We notice that the numerator is a quadratic equation. Is roots are :
x = -2 and x" = 10
Then the numerator could be written (x+2)(x+10) and the function:
y= (x+2)(x+10)/(x+2). Prior to simplifying, you should notice that this function is discontinued for x = -2 (doesn't exist), so the domain is all x# -2.
Now you simplify and you will get:
y = (x-10) , which is a linear function, but cannot be at x = -2
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tamaranim1 [39]

ANSWER

The required equation is:

9 {x}^{2}  - 25{y}^{2}  + 250y  - 85 0=0

EXPLANATION

The given equation is

9 {x}^{2}  - 25 {y}^{2}  = 225

Dividing through by 225 we obtain;

\frac{ {x}^{2} }{25}  -  \frac{ {y}^{2} }{9}  = 1

This is a hyperbola that has it's centre at the origin.

If this hyperbola is translated so that its center is now at (0,5).

Then its equation becomes:

\frac{ {(x - 0)}^{2} }{25}  -  \frac{ {(y - 5)}^{2} }{9}  = 1

We multiply through by 225 to get;

9 {x}^{2}  - 25( {y - 5})^{2}  = 225

We now expand to get;

9 {x}^{2}  - 25( {y}^{2} - 10y + 25 )= 225

9 {x}^{2}  - 25{y}^{2}  + 250y  - 6 25 = 225

The equation of the hyperbola in general form is

9 {x}^{2}  - 25{y}^{2}  + 250y  - 85 0=0

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