Question

The vertex of this parabola is at (-4, -1). When the y-value is 0, the x-value is 2. What is the coefficient of the squared term in the parabola's equation?

Answer 1

This is a parabola in vertex form,

`x = a(y - p)^2 + q`

a is the coefficient on y^2 we seek. Since it's sideways we have p=-1, q=-4

`x = a(y -1)^2 - 4`

When y=0,

`x=a - 4 = 2`

`a = 6`

Answer: A

Answer 2

Answer: Option A

Explanation:

The shape of the vertex of a parabola is:

`x = a (y-h) ^ 2 + k`

Where the point (k, h) is the vertex of the parabola and "a" is the coefficient of the squared term.

In this case we know that the vertex of this parabola is:

(-4, -1)

Then:

`h=-1` and
`k=-4`

So the equation is:

`x = a (y+1) ^ 2 -4`

We know that whe the y-value is 0, the x-value is 2.

This mean that:

`2 = a (0+1) ^ 2 -4`

`2 = a -4`

`a= 6`