Answer
The statement is False. The maximum value of the function is $\frac{5}{4}$.
Work Step by Step
The general form of a quadratic function is defined as $f\left( x \right)=a{{x}^{2}}+bx+c$.
When $a<0,$ the maximum value exists.
In order to find the maximum value, calculate $x=-\frac{b}{2a}.$
Putting in the value of a, b in $x=-\frac{b}{2a}$:
Thus,
$\begin{align}
& x=-\frac{1}{2\left( -1 \right)} \\
& =\frac{1}{2}
\end{align}$
Therefore, at $x=\frac{1}{2}$ the value of the function is a maximum.
Calculate the maximum value of the function as:
$\begin{align}
& f\left( \frac{1}{2} \right)=-{{\left( \frac{1}{2} \right)}^{2}}+\frac{1}{2}+1 \\
& =-\frac{1}{4}+\frac{1}{2}+1 \\
& =\frac{-1+2+4}{4} \\
& =\frac{5}{4}
\end{align}$
So, the maximum value of the quadratic function is $\frac{5}{4}$ at $x=\frac{1}{2}$.
Hence, the provided statement is false.
