Answer
a) See graph
b) Domain $(-\infty, \infty)$
Range$ [\frac{-10}{3}, \infty)$
c) Decreasing $(-\infty, \frac{4}{3}]$
Increasing $[\frac{4}{3}, \infty)$
Work Step by Step
$f(x) = 3x^2 - 8x + 2 $
a) $a = 3$, $b = -8$, $c = 2$
a > 0, so graph will open upwards
vertex $x = \frac{-b}{2a} = \frac{4}{3}$
Axis of symmetry $x = \frac{4}{3}$
Minimum value = $f(\frac{4}{3}) = 3(\frac{4}{3})^2 - 8\times (\frac{4}{3}) + 2 = \frac{-10}{3}$
For x intercept $y = 0$
we get $3x^2 - 8x + 2 = 0$
$=> x = \frac{4 \pm √\sqrt{10}}{3}$
y-intercept at $x = 0$, $(0, 2)$
b) Domain $(-\infty, \infty)$
Range$ [\frac{-10}{3}, \infty)$
c) Decreasing $(-\infty, \frac{4}{3}]$
Increasing $[\frac{4}{3}, \infty)$
