When x=-1:

Ok that gives us a little more information.
If we implicitly differentiate with respect to t, from the very start, then we can apply our product rule, ya?

The right side is zero, derivative of a constant is zero.
Where x' is dx/dt and y' is dy/dt.
From here, plug in all the stuff you know:
y' = -3
x = -1
y = 4
and solve for x'.
Hope that helps!
Answer:
x = 32
Step-by-step explanation:
Answer:
D. Minimum at (3, 7)
Step-by-step explanation:
We can add and subtract the square of half the x-coefficient:
y = x^2 -6x +(-6/2)^2 +16 -(-6/2)^2
y = (x -3)^2 +7 . . . . . simplify to vertex form
Comparing this to the vertex for for vertex (h, k) ...
y = (x -h)^2 +k
We find the vertex to be ...
(3, 7) . . . . vertex
The coefficient of x^2 is positive (+1), so the parabola opens upward and the vertex is a minimum.
Answer:
yes
Step-by-step explanation:
Jim's time is 5:50 and John's time is 6:40.