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Give the properties for the equation -2x 2 - y + 10x - 7 = 0.

>Vertex

(5/2, 11/2)
(-5/2, 11/2)
(5, -7)


Sagot :

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.