f(x) = x^3 − 8x^2. Find the point(s) on the graph of f where the tangent line is horizontal.?

F(x) = y = x³ - 8x²

Slope of tangent
= F ’(x)
= (d/dx)(x³ - 8x²)
= 3x² - 16x
= x(3x - 16)

For a horizontal line, slope = 0
x(3x - 16) = 0
x = 0 or x = 16/3

When x = 0 :
y = (0)³ - 8(0)²
y = 0

When x = 16/3 :
y = (16/3)³ - 8(16/3)²
y = -2048/27

The points are (0, 0) and (16/3, -2048/27)

Get the derivative, f'(x).
Then set f'(x) to zero. Then find x.
The x value will be the x-coordinate of the point where slope is horizontal.
Substitute x at the original equation and solve for f(x), and you will get the y-coordinate.

By definition, f'(x) is the slope of the curve. By setting it to zero, we will get the horizontal slope.

Hope this helps.

f ` (x) = 3x² - 16 x = 0
x [ 3x - 16 ] = 0
x = 0 , x = 16/3
f (0) = 0
f (16/3) = 4096/27 - 8 (256/9)
f (16/3) = 4096/27 - 24 (256/27)
f (16/3) = 4096 / 27 - 6144 / 27
f (16/3) = - 2048 / 27

Points are (0,0) and (16/3 - 2048/27)