Answer
a. $3(x+\frac{1}{3})^{2} - \frac{7}{3}$
b.
$Vertex: (\frac{-1}{3},\frac{-7}{3})$
$x-intercepts: \frac{-1 +\sqrt 7}{3}$ and $x= \frac{-1 -\sqrt 7}{3}$
$y-intercept: -2$
c. See the graph
d.
Domain: All real numbers
Range: $[\frac{-7}{3}, infinity)$
Work Step by Step
a.
$3x^{2} + 2x - 2 = 3(x^{2} + \frac{2}{3}x + \frac{1}{9}) – \frac{1}{9}\times3 - 2 = 3(x+\frac{1}{3})^{2} - \frac{7}{3}$
b.
$Vertex: (\frac{-1}{3},\frac{-7}{3})$
$x-intercepts: f(x)=0$
$3x^{2} + 2x - 2 = 0$
=> $x= \frac{-2 +\sqrt (2^{2}-4\times3\times(-2))}{2\times3}$ and $x= \frac{-2 -\sqrt (2^{2}-4\times3\times(-2))}{2\times3}$
=> $x= \frac{-2 +\sqrt (4+24)}{6}$ and $x= \frac{-2 -\sqrt (4+24)}{6}$
=> $x= \frac{-1 +\sqrt 7}{3}$ and $x= \frac{-1 -\sqrt 7}{3}$
$y-intercept: f(0) = 3\times0^{2} + 2\times0 - 2 = -2$
c. Sketch like in the graph
d.
Domain is all real numbers
Range is from $[\frac{-7}{3}, infinity)$
