Question

What is the equation of the parabola in vertex form x^2+6x-y+3=0?

Answer 1

Answer:

(x+3)^2=(y+6)

Step-by-step explanation:

the vertex of the parabola is (-3,-6) so i im just taking a guess based upon this assumption.

Answer 2

The vertex form of the equation of the parabola is $y = (x+3)^2 -6$.

What is the equation of the parabola?

The set of all points in a plane that are equidistant from a fixed line and a fixed point (not on the fixed line) in the plane is a parabola. The fixed line is the ‘directrix’ and the fixed point is the ‘focus’. The point of intersection of the parabola with the axis is the vertex.

For the given situation,

The equation is x^2+6x-y+3=0

The vertex form of the parabola is $y = (x-h)^2 + k$,

where the vertex is the point (h,k).

Rewrite the equation in the form $y = ax^2 + bx + c$

⇒ $-y=-x^{2} -6x-3$

Divide by -1 on both sides,

⇒ $y=x^{2} +6x+3$

Here a = 1, b = 6, c = 3

The axis of symmetry of the x-coordinate of the vertex is

$h = \frac{-b}{2a}$

⇒ $h = \frac{-6}{2(1)}$

⇒ $h=-3$

The y-coordinate of the vertex can be found by substitute the x value into the function,

⇒ $k=(-3)^{2} +6(-3)+3$

⇒ $k=9-18+3$

⇒ $k=-6$

So the vertex is (-3, -6).

Thus the vertex form of the parabola is $y = (x-h)^2 + k$,

⇒ $y = (x-(-3))^2 -6$

⇒ $y = (x+3)^2 -6$

Hence we can conclude that the vertex form of the equation of the parabola is $y = (x+3)^2 -6$.

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