For what values of x does the curve y^2-x^3-15x^2=8 have horizontal tangent lines?

-10 and 0.

Take the 1st derivative and set it = 0.
You completed the differentiation wisely and equating it to 0 is right to be certain the x fee of a horizontal tangent. 3x^2 – 8x – 7 = 0 Now we can use the quadratic equation: (-b ± sqrt(b^2 – 4ac)) / 2a the place a = 3, b = -8, c = -7 (8 ± sqrt(sixty 4 – 4 * 3 * -7)) / 6 (8 ± sqrt(148)) / 6 Now 148 = 4 * 37 which permits us to get rid of the 4 from the sqrt (sqrt(4) = 2) so (8 ± 2 * sqrt(37)) / 6 Divide the completed equation via 2 supplies (4 ± sqrt(37)) / 3
find when dy/dx = 0

y^2 – x^3 – 15x^2 = 8

2y * dy – 3x^2 * dx – 30x * dx = 0

dy * (2y) – dx * (3x^2 + 30x) = 0

dy * 2y = dx * (3x^2 + 30x)

dy/dx = (3x^2 + 30x) / (2y)

3x^2 + 30x = 0

x^2 + 10x = 0

x * (x + 10) = 0

x = -10 , 0