## College Algebra (10th Edition)

Write $f$ in the form $f(x)=a(x-h)^{2}+k$: \displaystyle \begin{aligned}f(x)&=\displaystyle \frac{2}{3}x^{2}+\frac{4}{3}x-1\\&=\displaystyle \frac{2}{3}\left(x^{2}+2x\right)-1\\&=\displaystyle \frac{2}{3}\left(x^{2}+2x+1\right)-1-\frac{2}{3}\\&=\displaystyle \frac{2}{3}(x+1)^{2}-\frac{5}{3}\end{aligned} Let $f_{1}(x)=x^{2}$. Then, $f(x)=\displaystyle \frac{2}{3}f_{1}(x+1)-\frac{5}{3}.$ The graph is obtained from $f_{1}(x)$ by - vertically compressing by factor $\displaystyle \frac{2}{3} \displaystyle \quad (x,y)\rightarrow(x,\frac{2}{3}y)$ - shifting to the left by $1$ units,$\quad \rightarrow (x-1,\displaystyle \frac{2}{3}y)$ - and then down by $\displaystyle \frac{5}{3}$ units $\displaystyle \quad \rightarrow(x-1,\frac{2}{3}y-\frac{5}{3})$ Point by point, $\left[\begin{array}{lll} (x,y) & \rightarrow & (x-1,\frac{2}{3}y-\frac{5}{3})\\ & & \\ (0,0) & \rightarrow & (-1,-\frac{5}{3})\\ (-1,1) & \rightarrow & (-2,-1)\\ (1,1) & \rightarrow & (0,-1)\\ (-2,4) & \rightarrow & (-3,1)\\ (2,4) & \rightarrow & (1,1) \end{array}\right]$