Answer:
[tex]g(n)=(n-10)^2-5[/tex]
The vertex will be at [tex](10, -5)[/tex]
[tex]x=10[/tex]
Step-by-step explanation:
The given quadratic function is,
[tex]g(n) = n^2-20n+95[/tex]
[tex]=n^2-2\cdot n\cdot 10+95[/tex]
[tex]=(n^2-2\cdot n\cdot 10+10^2)-10^2+95[/tex]
[tex]=(n-10)^2-100+95[/tex]
[tex]=(n-10)^2-5[/tex]
The vertex form is,
[tex]g(n)=(n-10)^2-5[/tex]
The vertex will be at [tex](10, -5)[/tex]
As the leading coefficient of [tex]g(n) = n^2-20n+95[/tex] is positive, so the parabola will open upwards. Hence, at the vertex the value will be minimum.
The axis of symmetry will be,
[tex]x=-\dfrac{b}{2a}[/tex]
Putting the values,
[tex]x=-\dfrac{-20}{2\times 1}=10[/tex]
