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(B) Given the parametric equations: $x = t^3 - 4t^2 - 3t$ $y = 2t^2 + 3t - 5$ First,find the derivatives with respect to $t$: $\frac{dx}{dt} = 3t^2 -

Vedclass · Accepted Answer

$H = 1$ and $V = 2$

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(B) Given the parametric equations: $x = t^3 - 4t^2 - 3t$ $y = 2t^2 + 3t - 5$ First,find the derivatives with respect to $t$: $\frac{dx}{dt} = 3t^2 - 8t - 3$ $\frac{dy}{dt} = 4t + 3$ The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t + 3}{3t^2 - 8t - 3}$. For a horizontal tangent,$\frac{dy}{dx} = 0$,which implies $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} 
eq 0$: $4t + 3 = 0 \implies t = -\frac{3}{4}$. At $t = -\frac{3}{4}$,$\frac{dx}{dt} = 3(-\frac{3}{4})^2 - 8(-\frac{3}{4}) - 3 = 3(\frac{9}{16}) + 6 - 3 = \frac{27}{16} + 3 
eq 0$. Thus,there is $1$ horizontal tangent,so $H = 1$. For a vertical tangent,$\frac{dx}{dt} = 0$ and $\frac{dy}{dt} 
eq 0$: $3t^2 - 8t - 3 = 0$ $(3t + 1)(t - 3) = 0$ $t = -\frac{1}{3}$ or $t = 3$. At these values,$\frac{dy}{dt} = 4t + 3 
eq 0$. Thus,there are $2$ vertical tangents,so $V = 2$. Therefore,$H = 1$ and $V = 2$.

Consider the curve represented parametrically by the equations $x = t^3 - 4t^2 - 3t$ and $y = 2t^2 + 3t - 5$,where $t \in \mathbb{R}$. If $H$ denotes the number of points on the curve where the tangent is horizontal and $V$ denotes the number of points where the tangent is vertical,then:

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