Answer:
[tex]b=20[/tex]
Step-by-step explanation:
Given:
The quadratic function is [tex]f(x)=-x^{2}+bx-75[/tex]
The maximum value of the function is 25.
Comparing it with the standard form, [tex]f(x)=ax^{2}+bx+c[/tex], we get
[tex]a=-1,b=b,c=-75[/tex]
Since, [tex]a[/tex] is negative, we have a downward parabola with maximum value at the vertex.
The vertex of a quadratic function occurs at [tex](h,k)=(\frac{-b}{2a},f(\frac{-b}{2a}))[/tex]
Now, [tex]h=\frac{-b}{2a}=\frac{-b}{-2}=\frac{b}{2}[/tex]
As per question, [tex]f(\frac{b}{2}) = 25[/tex]. This gives,
[tex]f(\frac{b}{2})=-(\frac{b}{2})^{2}+b(\frac{b}{2})-75\\25=-\frac{b^{2}}{4}+\frac{b^{2}}{2}-75\\25+75=\frac{b^{2}}{4}\\100=\frac{b^{2}}{4}\\b^{2}=100\times 4\\b=\sqrt{400}=20[/tex]
Therefore, the value of [tex]b[/tex] is 20.
