The maximum or minimum of a quadratic function \(\displaystyle{y}={a}{x}^{{{2}}}+{b}{x}+{c}\) occurs at the vertex which has an x-coordinate given by:

\(\displaystyle{x}=-{\frac{{{b}}}{{{2}{a}}}}\)

From the given, a=-1 and b=b sp we have:

\(\displaystyle{x}=-{\frac{{{b}}}{{{2}{\left(-{1}\right)}}}}={\frac{{{b}}}{{{2}}}}\)

Since a=-1

\(\displaystyle-{\left({\frac{{{b}}}{{{2}}}}\right)}^{{{2}}}+{b}{\left({\frac{{{b}}}{{{2}}}}\right)}-{75}={25}\)

Solve for b:

\(\displaystyle-{\frac{{{b}^{{{2}}}}}{{{4}}}}+{\frac{{{b}^{{{2}}}}}{{{2}}}}-{75}={25}\)

\(\displaystyle{\frac{{{b}^{{{2}}}}}{{{4}}}}={100}\)

\(\displaystyle{b}^{{{2}}}={400}\)

\(\displaystyle{b}=\pm\sqrt{{{400}}}\)

\(\displaystyle{b}=\pm{20}\)

Result:

\(\displaystyle{b}=\pm{20}\)

\(\displaystyle{x}=-{\frac{{{b}}}{{{2}{a}}}}\)

From the given, a=-1 and b=b sp we have:

\(\displaystyle{x}=-{\frac{{{b}}}{{{2}{\left(-{1}\right)}}}}={\frac{{{b}}}{{{2}}}}\)

Since a=-1

\(\displaystyle-{\left({\frac{{{b}}}{{{2}}}}\right)}^{{{2}}}+{b}{\left({\frac{{{b}}}{{{2}}}}\right)}-{75}={25}\)

Solve for b:

\(\displaystyle-{\frac{{{b}^{{{2}}}}}{{{4}}}}+{\frac{{{b}^{{{2}}}}}{{{2}}}}-{75}={25}\)

\(\displaystyle{\frac{{{b}^{{{2}}}}}{{{4}}}}={100}\)

\(\displaystyle{b}^{{{2}}}={400}\)

\(\displaystyle{b}=\pm\sqrt{{{400}}}\)

\(\displaystyle{b}=\pm{20}\)

Result:

\(\displaystyle{b}=\pm{20}\)