Answer
The axis of symmetry is $x=\displaystyle \frac{7}{4}$.
domain: $(-\infty,\infty)$
range: $[-\displaystyle \frac{81}{8},\infty)$
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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$f(x)=2x^{2}-7x-4$
$f(x)=2(x^{2}-\displaystyle \frac{7}{2}x+\frac{49}{16})-4-\frac{49}{8}$
$f(x)=2(x-\displaystyle \frac{7}{4})^{2}-\frac{81}{8}$
1. opens up (a=$2>0$).
2. vertex: $(\displaystyle \frac{7}{4},-\frac{81}{8}). $The axis of symmetry is $x=\displaystyle \frac{7}{4}.$
3. x-intercepts:
$2(x-\displaystyle \frac{7}{4})^{2}-\frac{81}{8}=0$
$2(x-\displaystyle \frac{7}{4})^{2}=\frac{81}{8}$
$(x-\displaystyle \frac{7}{4})^{2}=\frac{81}{16}$
$x-\displaystyle \frac{7}{4}=\pm\frac{9}{4}$
$x=\displaystyle \frac{7}{4}\pm\frac{9}{4}$
$x=-\displaystyle \frac{1}{2}$ ,$\quad x=4$
4. y-intercept:
$f(0)=2(0)^{2}-7(0)-4=-4$
The axis of symmetry is $x=\displaystyle \frac{7}{4}$.
domain: $(-\infty,\infty)$
range: $[-\displaystyle \frac{81}{8},\infty)$
