Question

Determining the Vertex of a Parabola from Its Graph

On a coordinate plane, a parabola opens up with solid circles along the parabola at (negative 6, 5), (negative 5, 0), (negative 4, negative 3), (negative 3, negative 4), (negative 2, negative 3), (negative 1, 0), (0, 5).

What is the vertex of the parabola in the graph?



(
,
)

Answer 1

(-3, -4)

Explanation:

The solid circles along the parabola are:

(-6, 5), (-5, 0), (-4, -3), (-3, -4), (-2, -3), (-1, 0), (0, 5).

Although on observation, the minimum point of y occurs at (-3, -4), we can also confirm through the function.

The x-intercept of the parabola are -5 and -1.

x=-5 or x=-1

x+5=0 or x+1=0

(x+5)(x+1)=0

`x^2+5x+x+5=0\\x^2+6x+5=0`

The Vertex of the equation occurs at the axis of symmetry,
`x=-(b)/(2a)`

In
`f(x)=x^2+6x+5, a=1, b=6`

Axis of Symmetry,
`x=-(6)/(2)=-3`

`f(-3)=(-3)^2+6(-3)+5=9-18+5=-4`

Therefore, we can confirm that the vertex is (-3,-4) as stated earlier.