Bayangan kurva y= x2+3x+3 jika dicerminkan terhadap sumbu X dilanjutkan dengan dilatasi pusat O dan faktor skala 3 adalah. terima kasih

jawab

y = x² + 3x + 3

T1= Rx
T2= D[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]= \left[\begin{array}{ccc}3x\\3y\end{array}\right] [/tex]

x' = 3x  → x =  1/3 x'
y'= - 3y → y = - 1/3 y'
subke  y = x² + 3 x + 3

(-1/3 y') = (1/3 x')² + 3 (1/3 x') +  3
- 1/3 y = 1/9x²  + x + 3 .....kali  (-3
y = 1/3 x² - 3x - 9