2016-I-18 – Solving Master

Ans: (a)

(18, - 13)

(b)

g (x) = \frac{- 1}{3} (x - 18)^{2}

y

-axis

$\begin{array}{rcl} y & = & f (x) \\ = & \frac{- 1}{3} x^{2} + 12 x - 121 \\ = & \frac{- 1}{3} (x^{2} - 36 x) - 121 \\ = & \frac{- 1}{3} (x^{2} - 36 x + 18^{2} - 18^{2}) - 121 \\ = & \frac{- 1}{3} (x - 18)^{2} + 108 - 121 \\ = & \frac{- 1}{3} (x - 18)^{2} - 13 \end{array}$

Therefore, the coordinates of the vertex of the graph of $y = f (x)$ are $(18, - 13)$ .
From the result of (a), the coordinates of the vertex of the graph of $y = f (x)$ are $(18, - 13)$ . Hence, to obtain the graph of $y = g (x)$ by translating the graph of $y = f (x)$ vertically, we have to translate the graph of $y = f (x)$ upwards $13$ units. Therefore, we have $\begin{array}{rcl} y & = & g (x) \\ = & f (x) + 13 \\ = & (\frac{- 1}{3} (x - 18)^{2} - 13) + 13 \\ = & \frac{- 1}{3} (x - 18)^{2} \end{array}$
Note that $\begin{array}{rcl} f (- x) & = & \frac{- 1}{3} (- x)^{2} + 12 (- x) - 121 \\ = & \frac{- 1}{3} x^{2} - 12 x - 121 \end{array}$

Hence, the mentioned transformation is reflecting $y = f (x)$ about $y$ -axis.