y - 3
g(y) = ------------------
y^2 - 3y + 9
To find the c. v., we must differentiate this function g(y) and set the derivative equal to zero:
(y^2 - 3y + 9)(1) - (y - 3)(2y - 3)
g '(y) = --------------------------------------------
(y^2 - 3y + 9)^2
Note carefully: The denom. has no real roots, so division by zero is not going to be an issue here.
Simplifying the denominator of the derivative,
(y^2 - 3y + 9)(1) - (y - 3)(2y - 3) => y^2 - 3y + 9 - [2y^2 - 3y - 6y + 9], or
-y^2 + 6y
Setting this result = to 0 produces the equation y(-y + 6) = 0, so
y = 0 and y = 6. These are your critical values. You may or may not have max or min at one or the other.
So you will use slope formula which is m = (y2 -- y1) / (x2 -- x1)
-1 minus -7=6
13 minus -8=21
Remember it changes to a positive Keep,Change,Change
So its 6/21
Or 1/4
Ask questions if you still need help
:)
A is your answer. Parallel lines never intersect. They both run the same direction, the same distance from each other.
_________________________
_________________________
are parallel lines
Do 6.5” divided by 5 to find the slope . Then plot each point and
