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Furkat [3]
3 years ago
11

Help please I need 5 points total. i need 2 to left of vertex. i need vertex. i need 2 to right of vertex. Please, quickly, I am

timed
graph only goes up to 14. thank you​

Mathematics
1 answer:
zlopas [31]3 years ago
5 0

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 ;

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

Where , D = b² - 4ac (Discriminant)

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

⇢⇢⇢ <em><u>a = 1 , b = - 10 , c = 2</u></em><em><u>7</u></em>

Now , Calculating D ;

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

{:\implies \quad \sf D=100-108}

{:\implies \quad \bf \therefore \quad D=-8}

Now , Calculating the vertex ;

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

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

{:\implies \quad \bf \therefore \quad Vertex = (5,2)}

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 .

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