Question

Which of the following equations is of a parabola with a vertex at (0, -5)?

1. y = x2 - 5
2. y = x2 + 5
3. y = ( x + 5) 2
4. y = ( x - 5) 2

Answer 1

The equation that has a vertex in (0, -5) is the first one.

Explanation:

A standard form second degree equation is given by the following expression:

`y(x) = a*x^2 + b*x + c`

For which the vertex coordinates can be calculated by:

`x_(vertex) = -(b)/(2*a)`

While "y" for the vertex can be found by applying this coordinate on the expression. Using this knowledge in each equation gives us:

1.
`y(x) = x^2 - 5`

`a = 1\\b = 0\\c = -5`

Therefore the vertex coordinate is:

`x_(vertex) = -(b)/(2*a) = -(0)/(2*1) = 0`

`y_(vertex) = 0^2 - 5 = -5`

This parabola has a vertex in (0,-5).

2.
`y(x) = x^2 + 5`

`a = 1\\b = 0\\c = 5`

Therefore the vertex coordinate is:

`x_(vertex) = -(b)/(2*a) = -(0)/(2*1) = 0`

`y_(vertex) = 0^2 + 5 = 5`

This parabola has a vertex in (0,5).

3.
`y(x) =(x+ 5)^2 = x^2 + 10*x + 25`

`a = 1\\b = 10\\c = 25`

Therefore the vertex coordinate is:

`x_(vertex) = -(b)/(2*a) = -(10)/(2*1) = -5`

`y_(vertex) = (-5)^2 + 10*(-5) + 25 = 0`

This parabola has a vertex in (-5,0).

4.
`y(x) =(x- 5)^2 = x^2 - 10*x + 25`

`a = 1\\b =- 10\\c = 25`

Therefore the vertex coordinate is:

`x_(vertex) = -(b)/(2*a) = -(-10)/(2*1) = 5`

`y_(vertex) = (5)^2 - 10*(5) + 25 = 0`

This parabola has a vertex in (5,0).