To find the normal slope at (2,−1), first find the value of t that gives this point. Solve: t2+3t−8=2⇒t2+3t−10=0 This factors: (t+5)(t−2)=0⇒t=2‌ (since it must give ‌y=−1) Now compute derivatives: ‌
dx
dt
=2t+3,‌‌‌
dy
dt
=4t−2. At t=2 : ‌
dx
dt
=7,‌‌‌
dy
dt
=6 So the slope of the tangent is: ‌
dy
dx
=‌
6
7
Thus, the slope of the normal is the negative reciprocal: m‌normal ‌=−‌