Answer
The graph is shown below.
Range $\left(-\infty,\frac{5}{4}\right]$.
Work Step by Step
Rearrange the given function.
$f(x)=-(x-\frac{1}{2})^2+\frac{5}{4}$
The standard form of the parabola is
$f(x)=a(x-h)^2+k$
Compare with the original given function.
$a=-1,h=\frac{1}{2}$ and $k=\frac{5}{4}$
Step 1:- Parabola opens.
$a<0$, The parabola opens downward.
Step 2:- Vertex.
The value of $h=\frac{1}{2}$ and $k=\frac{5}{4}$. The vertex is $(h,k)=(\frac{1}{2},\frac{5}{4})$.
Step 3:- $x-$intercepts.
Replace $f(x)$ with $0$.
$\Rightarrow 0=-(x-\frac{1}{2})^2+\frac{5}{4}$
Add $(x-\frac{1}{2})^2$ to both sides.
$\Rightarrow 0+(x-\frac{1}{2})^2=-(x-\frac{1}{2})^2+\frac{5}{4}+(x-\frac{1}{2})^2$
Simplify.
$\Rightarrow (x-\frac{1}{2})^2=\frac{5}{4}$
Apply the square root property.
$\Rightarrow x-\frac{1}{2}=\sqrt{\frac{5}{2^2}}$ or $x-\frac{1}{2}=-\sqrt{\frac{5}{2^2}}$
Add $\frac{1}{2}$ to both sides in each equation.
$\Rightarrow x-\frac{1}{2}+\frac{1}{2}=\frac{\sqrt{5}}{2}+\frac{1}{2}$ or $x-\frac{1}{2}+\frac{1}{2}=-\frac{\sqrt{5}}{2}+\frac{1}{2}$
Simplify.
$\Rightarrow x=\frac{\sqrt{5}+1}{2}$ or $x=\frac{-\sqrt{5}+1}{2}$
Hence, the $x-$intercepts are $\frac{\sqrt{5}+1}{2}$ and $\frac{-\sqrt{5}+1}{2}$. The parabola passes through $(\frac{\sqrt{5}+1}{2},0)$ and $(\frac{-\sqrt{5}+1}{2},0)$.
Step 4:- $y-$intercept.
Replace $x$ with $0$.
$\Rightarrow f(0)=-(0-\frac{1}{2})^2+\frac{5}{4}$
Simplify.
$\Rightarrow f(0)=-\frac{1}{4}+\frac{5}{4}$
Simplify.
$\Rightarrow f(0)=\frac{-1+5}{4}$
$\Rightarrow f(0)=\frac{4}{4}$
$\Rightarrow f(0)=1$
Hence, the $y-$intercept is $1$. The parabola passes through $(0,1)$.
Step 5:- Graph.
Use the points vertex, $x-$intercepts and $y-$intercept to draw a parabola.
The axis of symmetry is $x=h=\frac{1}{2}$.
$A=(\frac{1}{2},\frac{5}{4})$
$B=(\frac{\sqrt{5}+1}{2},0)$
$C=(\frac{-\sqrt{5}+1}{2},0)$
$D=(0,1)$.
From the graph the range of the function is
$\left(-\infty,\frac{5}{4}\right]$.
