Given:
The quadratic equation is
[tex]-2x^2-y+10x-7=0[/tex]
To find:
The vertex of the given quadratic equation.
Solution:
If a quadratic function is [tex]f(x)=ax^2+bx+c[/tex], then
[tex]Vertex=\left(\dfrac{-b}{2a},f(\dfrac{-b}{2a})\right)[/tex]
We have,
[tex]-2x^2-y+10x-7=0[/tex]
It can be written as
[tex]-2x^2+10x-7=y[/tex]
[tex]y=-2x^2+10x-7[/tex] ...(i)
Here, [tex]a=-2,b=10,c=-7[/tex].
[tex]\dfrac{-b}{2a}=\dfrac{-10}{2(-2)}[/tex]
[tex]\dfrac{-b}{2a}=\dfrac{-10}{-4}[/tex]
[tex]\dfrac{-b}{2a}=\dfrac{5}{2}[/tex]
Putting [tex]x=\dfrac{5}{2}[/tex] in (i), we get
[tex]y=-2(\dfrac{5}{2})^2+10(\dfrac{5}{2})-7[/tex]
[tex]y=-2(\dfrac{25}{4})+\dfrac{50}{2}-7[/tex]
[tex]y=\dfrac{-50}{4}+25-7[/tex]
[tex]y=\dfrac{-25}{2}+18[/tex]
On further simplification, we get
[tex]y=\dfrac{-25+36}{2}[/tex]
[tex]y=\dfrac{11}{2}[/tex]
So, the vertex of the given quadratic equation is [tex]\left(\dfrac{5}{2},\dfrac{11}{2}\right)[/tex].
Therefore, the correct option is A.
