# Question #34c6f

Jul 19, 2016

horizontal tangents $t = \pm 2$ .

vertical tangents $t = 1$

#### Explanation:

well $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

$= \frac{3 {t}^{2} - 12}{2 t - 2}$

you will have horizontal tangents when the numerator = 0

ie $3 {t}^{2} = 12 , t = \pm 2$ ..... and just check that he denominator is not also 0 at $t = \pm 2$, as that might cause problems

you will have vertical tangents when the denominator = 0

ie $2 t = 2 , t = 1$