Answer
Domain of $f$ is $(-∞, ∞)$
Range of $f$ is $[-2,∞)$
Work Step by Step
$f(x)=3x^2-6x+1$
$a=3$, $b=-6$, $c=1$
Graphing:
The coefficient of $x$ is $3$, therefore the parabola opens upward.
Get the vertex: $x=\frac{-b}{2a}$
$x=\frac{-(-6)}{2(3)}$
$x=1$
$f(1)=3(1)^2-6(1)+1$
$f(1)=-2$
vertex is at $(1,-2)$
Find the $x$-intercepts by solving $f (x) = 0$.
$0=3x^2-6x+1$
Use the quadratic formula to find the values of $x$.
$x = \frac{-b±\sqrt{b^2-4ac}}{2a}$
$a=3$, $b=-6$, $c=1$
$x = \frac{-(-6)±\sqrt{-6^2-(4⋅3⋅1)}}{2⋅3}$
$x = \frac{6±\sqrt{36-12}}{6}$
$x = \frac{6±\sqrt{24}}{6}$
$x = \frac{6±\sqrt{(4⋅6)}}{6}$
$x = \frac{6±2\sqrt{6}}{6}$
$x = \frac{3±\sqrt{6}}{3}$
$x=\frac{3+\sqrt{6}}{3}$ or $x=\frac{3-\sqrt{6}}{3}$
The $x$-intercepts are at points $(\frac{3+\sqrt{6}}{3},0)$ and $(\frac{3-\sqrt{6}}{3},0)$
Domain: The graph widens and continues to fall at both ends, and thus, include all real numbers
Domain of $f$ is $(-∞, ∞)$
Range: The parabola’s vertex, (1, -2), is the lowest point on the graph. Because the y-coordinate of the vertex is $-2$, outputs on the y-axis fall at or above $-2$.
Range of $f$ is $[-2,∞)$
