Answer
The graph is shown below.
Range $\left[-\frac{13}{3},\infty\right)$.
Work Step by Step
The given function is a quadratic function:
$f(x)=3x^2-2x-4$
The standard form of the quadratic function is
$f(x)=ax^2+bx+c$
Compare both equations $a=3,b=-2$ and $c=-4$.
Step 1:- Parabola opens.
$a>0$, the parabola open upward.
Step 2:- Vertex.
$x-$coordinate of the vertex is $x=-\frac{b}{2a}=-\frac{(-2)}{2(3)}=\frac{1}{3}$.
Substitute the value of $x$ into the function.
$\Rightarrow f(\frac{1}{3})=3(\frac{1}{3})^2-2(\frac{1}{3})-4$
Simplify.
$\Rightarrow f(\frac{1}{3})=\frac{1}{3}-\frac{2}{3}-4$
The LCD is $3$.
Multiply the numerator and the denominator to form LCD at the denominator.
$\Rightarrow f(\frac{1}{3})=\frac{1}{3}-\frac{2}{3}-\frac{12}{3}$
Add numerators because denominators are same.
$\Rightarrow f(\frac{1}{3})=\frac{1-2-12}{3}$
$\Rightarrow f(\frac{1}{3})=-\frac{13}{3}$
The vertex is $\left(\frac{1}{3},-\frac{13}{3}\right)$.
Step 3:- $x-$intercepts.
Replace $f(x)$ with $0$ into the given function.
$\Rightarrow 0=3x^2-2x-4$
Use quadratic formula, we have $a=3,b=-2$ and $c=-4$.
$\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Substitute all values.
$\Rightarrow x=\frac{-(-2)\pm\sqrt{(-2)^2-4(3)(-4)}}{2(3)}$
Simplify.
$\Rightarrow x=\frac{2\pm\sqrt{4+48}}{6}$
$\Rightarrow x=\frac{2\pm\sqrt{52}}{6}$
$\Rightarrow x=\frac{2\pm2\sqrt{13}}{6}$
Separate the fractions.
$\Rightarrow x=\frac{2+2\sqrt{13}}{6}$ or $x= \frac{2-2\sqrt{13}}{6}$
Simplify.
$\Rightarrow x=\frac{1+\sqrt{13}}{3}$ or $x= \frac{1-\sqrt{13}}{3}$
The $x-$intercepts are $\frac{1+\sqrt{13}}{3}$ and $\frac{1-\sqrt{13}}{3}$. The parabola passes through $\left(\frac{1+\sqrt{13}}{3},0\right)$ and $\left(\frac{1-\sqrt{13}}{3},0\right)$.
Step 4:- $y-$intercept.
Replace $x$ with $0$ in the given function.
$\Rightarrow f(0)=3(0)^2-2(0)-4$
Simplify.
$\Rightarrow f(0)=-4$
The $y-$intercept is $-14$. The parabola passes through $(0,-4)$.
Step 5:- Graph.
Use the points:- vertex, $x-$intercepts and $y-$intercept to draw a parabola.
The axis of symmetry is $x=\frac{1}{3}$.
$A=\left(\frac{1}{3},-\frac{13}{3}\right)$
$B=\left(\frac{1+\sqrt{13}}{3},0\right)$
$C=\left(\frac{1-\sqrt{13}}{3},0\right)$
$D=(0,-4)$.
From the graph the range of the function is
$\left[-\frac{13}{3},\infty\right)$.
