Answer
The axis of symmetry is $x=\displaystyle \frac{1}{3}.$
domain: $(-\infty,\infty)$
range: $[-\displaystyle \frac{13}{3},\infty)$
.
Work Step by Step
First, rewrite $f(x)=ax^{2}+bx+c$ in standard form, $f(x)=a(x-h)^{2}+k$
Graphing:
1. Determine whether the parabola opens upward or downward.
If $a>0$, it opens upward. If $a<0$, it opens downward.
2. Determine the vertex of the parabola. The vertex is $(h, k)$.
3. Find any x-intercepts by solving $f(x)=0$.
The function's real zeros are the x-intercepts.
4. Find the y-intercept by computing $f(0)$.
5. Plot the intercepts, the vertex, and additional points as necessary
Connect these points with a smooth curve that is shaped like a bowl or an inverted bowl.
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$f(x)=3x^{2}-2x-4$
$f(x)=3(x^{2}-\displaystyle \frac{2}{3}x )-4$
$f(x)=3(x^{2}-\displaystyle \frac{2}{3}x+\frac{1}{9})-4-\frac{1}{3}$
$f(x)=3(x-\displaystyle \frac{1}{3})^{2}-\frac{13}{3}$
1. opens up (a = $3 > 0$).
2. vertex: $(\displaystyle \frac{1}{3},-\frac{13}{3}). $The axis of symmetry is $x=\displaystyle \frac{1}{3}.$
3. x-intercepts:
$3(x-\displaystyle \frac{1}{3})^{2}-\frac{13}{3}=0$
$3(x-\displaystyle \frac{1}{3})^{2}=\frac{13}{3}$
$(x-\displaystyle \frac{1}{3})^{2}=\frac{13}{9}$
$x-\displaystyle \frac{1}{3}=\pm\sqrt{\frac{13}{9}}$
$x=\displaystyle \frac{1}{3}\pm\frac{\sqrt{13}}{3}$
4. y-intercept:
$f(0)=3(0)^{2}-2(0)-4=-4$
The axis of symmetry is $x=\displaystyle \frac{1}{3}.$
domain: $(-\infty,\infty)$
range: $[-\displaystyle \frac{13}{3},\infty)$
