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Sagot :
Given:
The equation of the parabola is
[tex]y=2x^2+8x+3[/tex]
To find:
The vertex and check whether it opens up or down.
Solution:
We have,
[tex]y=2x^2+8x+3[/tex]
Here, leading coefficient is 2 which is a positive number. So, the parabola opens up.
The vertex of parabola [tex]f(x)=ax^2+bx+c[/tex] is
[tex]\left(\dfrac{-b}{2a},f(\dfrac{-b}{2a})\right)[/tex]
Here, a=2, b=8 and c=3.
[tex]\dfrac{-b}{2a}=\dfrac{-8}{2(2)}[/tex]
[tex]\dfrac{-b}{2a}=\dfrac{-8}{4}[/tex]
[tex]\dfrac{-b}{2a}=-2[/tex]
Putting x=-2 in the given equation.
[tex]y=2(-2)^2+8(-2)+3[/tex]
[tex]y=2(4)+(-16)+3[/tex]
[tex]y=8-16+3[/tex]
[tex]y=-5[/tex]
So, the vertex of the parabola is at (-2,-5) and parabola opens up.
Therefore, the correct option is D.
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