Respuesta :

We are given with a quadratic equation which represents a Parabola , we need to find the vertex of the parabola , But let's recall that , For any quadratic equation of the form ax² + bx + c = 0 , the vertex of the parabola is given by ;

[tex]{\quad \qquad \boxed{\bf{Vertex = \left(\dfrac{-b}{2a},\dfrac{-D}{4a}\right)}}}[/tex]

Where , D = b² - 4ac (Discriminant)

Now , On comparing the given equation with ax² + bx + c , we have

a = 1 , b = - 10 , c = 27

Now , Calculating D ;

[tex]{:\implies \quad \sf D=(-10)^{2}-4\times 1\times 27}[/tex]

[tex]{:\implies \quad \sf D=100-108}[/tex]

[tex]{:\implies \quad \bf \therefore \quad D=-8}[/tex]

Now , Calculating the vertex ;

[tex]{:\implies \quad \sf Vertex =\bigg\{\dfrac{-(-10)}{2\times 1},\dfrac{-(-8)}{4\times 1}\bigg\}}[/tex]

[tex]{:\implies \quad \sf Vertex = \left(\dfrac{10}{2},\dfrac{8}{2}\right)}[/tex]

[tex]{:\implies \quad \bf \therefore \quad Vertex = (5,2)}[/tex]

Hence , The vertex of the parabola is at (5,2)

Note :- As the Discriminant < 0 . So , the equation will have imaginary roots .

Refer to the attachment for the graph as well .

Ver imagen Аноним
ACCESS MORE
EDU ACCESS
Universidad de Mexico