Answer:
There is no point of the form (-1, y) on the curve where the tangent is horizontal
Step-by-step explanation:
Notice that when x = - 1. then dy/dx becomes:
dy/dx= (y+2) / (2y+1)
therefore, to request that the tangent is horizontal we ask for the y values that make dy/dx equal to ZERO:
0 = ( y + 2) / (2 y + 1)
And we obtain y = -2 as the answer.
But if we try the point (-1, -2) in the original equation, we find that it DOESN'T belong to the curve because it doesn't satisfy the equation as shown below:
(-1)^2 + (-2)^2 - (-1)*(-2) - 5 = 1 + 4 + 2 - 5 = 2 (instead of zero)
Then, we conclude that there is no horizontal tangent to the curve for x = -1.
Answer:
Step-by-step explanation:
Combining like-terms gives us 
Hope this helped!
I'm probably wrong but I think A.
Answer: The result is one half (
)
Step-by-step explanation:
We have the following expression:
Since both fractions have the same denominator, we can just add both numerators and keep the denominator:
Dividing numerator and denominator by 
:
This is the result
