## Intermediate Algebra for College Students (7th Edition)

Domain of $f$ is $(-∞, ∞)$ Range of $f$ is $[-2,∞)$
$f(x)=3x^2-6x+1$ $a=3$, $b=-6$, $c=1$ 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 $[-2,∞)$