how do you find in which parabola opens also the vertex? y^2 - 10y - x + 22 = 0?

Question

also how do you do that with tranformation to y=..........?

Answer 1

y²-10y+25 - 25+22=x
(y-5)² - 3 = x
vertex is at (-3, 5)
since it is squared in the variable y it will open to the right rather than up.

(y-5)²=x+3
y-5= ±√(x+3)
y= 5±√(x+3)

Answer 2

parabolas that look like y-k = a(x-h)² open up or down (depending on the sign of a); they don't contain y². (h,k) is the vertex. equations in which y is squared but x is not open left or right:

y² - 10y - x + 22 = 0
y² - 10y = x - 22
y² - 10y + 25 = x - 22 + 25
(y - 5)² = x + 3
so this parabola opens right, vertex at (-3,5).

Answer 3

Since y² is positive, you can count on the parabola opening upward or to the right. In this case, it would be to the right.

You can use the Quadratic Formula on this, just like any other quatdratic:

y = [-b +/- sqrt(b² - 4ac)] / (2a)

In this formula, a = 1, b = -10, and c = (-x + 22) or c = -(x - 22).

Good luck!

Answer 4

bypass the 9 to the main appropriate factor, then finished the sq. in x and y on the left: x^2 + 2x + ( ) + y^2 + 4y + ( ) = 9 x^2 + 2x + a million + y^2 + 4y + 4 = 9 + a million + 4 (x + a million)^2 + (y + 2)^2 = 14 midsection is (-a million, -2), radius is sqrt(14)