Answer
$(-3,37)$ and $(1,5)$
Work Step by Step
Recall: The tangent of $y=f(x)$ is horizontal when $f'(x)=0$.
We have $f(x)=x^3+3x^2-9x+10$.
Find $f'(x)$:
$f'(x)=3x^{3-1}+3\cdot 2x^{2-1}-9+0$
$f'(x)=3x^2+6x-9$
Find the values of $x$ such that $f'(x)=0$:
$3x^2+6x-9=0$ (Divide by 3)
$x^2+2x-3=0$ (Factorize the left side)
$(x+3)(x-1)=0$
$x=-3\vee x=1$
For $x=-3$,
$y=(-3)^3+3\cdot (-3)^2-9\cdot (-3)+10$
$y=-27+27+27+10$
$y=37$
$\therefore (-3,37)$
For $x=1$
$y=1^3+3\cdot 1^2-9\cdot 1+10$
$y=1+3-9+10$
$y=5$
$\therefore (1,5)$
Thus, the points on the curve of $y=x^3+3x^2-9x+10$ where the tangent is horizontal are $(-3,37)$ and $(1,5)$.
