The slope of the tangent to the curve $x=t^2+3t-8$ ,$y=2t^2-2t-5$ at the point (2, –1) is

We need to determine the value of $t$ corresponding to the given point, so

$\left.2=t^2+3t-8\Rightarrow t^2+3t-10=0\right]\Rightarrow t=2,-5$

$-1=2t^2-2t-5\Rightarrow t^2-t-2=0 \Rightarrow t=2,-1$

Hence common value is 2

${\left.y^{\text{'}}\right|}_{x=2}={\left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right)}_{t=2}={\left(\frac{4t-2}{2t+3}\right)}_{t=2}=\frac{6}{7}$