persamaan peta kurva y = x² - 3x + 2 karena pencerminan terhadap sumbu x dilanjutkan dilatasi pusat 0 dengan faktor skalar 3 adalah

jawab

y = x² - 3x + 2
T1 pencerminan terhadap sumbu x
T2 dilatasi [O, 3]

[x',y'] = T2 O T1 [x, y]

[tex] \left[\begin{array}{ccc}x'\\y'\end{array}\right]\,= \left[\begin{array}{ccc}3&0\\0&3\end{array}\right] \left[\begin{array}{ccc}1&0\\0&-1\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc}x'\\y'\end{array}\right]\,= \left[\begin{array}{ccc}3&0\\0&-3\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right][/tex]

x' =  3x → x = 1/3 x'
y' = - 3y → y= - 1/3 y'
subs ke y = x² - 3x + 2
(-1/3 y') = (1/3 x')² - 3(-1/3 x')  + 2
- 1/3 y = 1/9x² + x + 2 ......kalikan -3
y = - 1/3 x² - 3x  - 18