Answer
See image
Work Step by Step
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,
$\left[\begin{array}{lll}
(x,y) & \rightarrow & (x-1,\frac{2}{3}y-\frac{5}{3})\\
& & \\
(0,0) & \rightarrow & (-1,-\frac{5}{3})\\
(-1,1) & \rightarrow & (-2,-1)\\
(1,1) & \rightarrow & (0,-1)\\
(-2,4) & \rightarrow & (-3,1)\\
(2,4) & \rightarrow & (1,1)
\end{array}\right]$
