Answer
The axis of symmetry is $x=\displaystyle \frac{1}{2}$.
domain: $(-\infty,\infty)$
range: $(-\infty,\displaystyle \frac{5}{4}]$
Work Step by Step
To graph $f(x)=a(x-h)^{2}+k$,
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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$f(x)=\displaystyle \frac{5}{4}-(x-\frac{1}{2})^{2}$
$f(x)=-(x-\displaystyle \frac{1}{2})^{2}+\frac{5}{4}$
1. opens down (a=$-1$).
2. vertex: $(\displaystyle \frac{1}{2},\frac{5}{4}). $The axis of symmetry is $x=\displaystyle \frac{1}{2}$.
3. x-intercepts:
$0=-(x-\displaystyle \frac{1}{2})^{2}+\frac{5}{4}$
$(x-\displaystyle \frac{1}{2})^{2}=\frac{5}{4}$
$x-\displaystyle \frac{1}{2}=\pm\frac{\sqrt{5}}{2}$
$x=\displaystyle \frac{1\pm\sqrt{5}}{2}$
4.y-intercept:
$f(0)=-\displaystyle \left(\begin{array}{l}
0-\underline{1}\\
2
\end{array}\right)+\frac{5}{4}=1$
5. see graph
The axis of symmetry is $x=\displaystyle \frac{1}{2}$.
domain: $(-\infty,\infty)$
range: $(-\infty,\displaystyle \frac{5}{4}]$
