Find the points on the graph y = -x^3 + 6x^2 at which the tangent line is horizontal.
I found -x^2 + 4 = 0
are the points [4,32]
y'=0=-3x^2+12x or
x(-x+4)=0
so the x points are 0,4
so the graph points are
(0,0) and (4,32)
To find the points on the graph where the tangent line is horizontal, we need to find the values of x where the derivative of the function is equal to zero. This is because the derivative represents the rate of change or slope of the function at any given point.
First, let's find the derivative of the function y = -x^3 + 6x^2. We can do this by taking the derivative of each term separately using the power rule:
dy/dx = -3x^2 + 12x
Now, set the derivative equal to zero and solve for x:
-3x^2 + 12x = 0
Next, factor out an x from the equation:
x(-3x + 12) = 0
Now, we have two possible solutions:
1) x = 0
2) -3x + 12 = 0
Solving the second equation, we get:
-3x = -12
x = 4
So, there are two possible values of x where the tangent line is horizontal: x = 0 and x = 4.
To find the corresponding y-values, substitute each value of x back into the original function:
For x = 0,
y = -0^3 + 6(0^2)
y = 0
For x = 4,
y = -4^3 + 6(4^2)
y = -64 + 96
y = 32
Therefore, the points on the graph where the tangent line is horizontal are (0, 0) and (4, 32).