Question

What is the distance between the y-intercepts of the graph of `x + 8 = 2(y+3)^2`?

Answer 1

Distance between y-intercepts is `4`.

Explanation:

Given equation is `x+8=2(y+3)^2`

One must know that when the function crosses the y-axis, the `x` parameter equals zero, that is `x=0`. So substituting for `x` in the equation, we get
`cancel{0+8}^4=cancel{2}^1(y+3)^2implies4=(y+3)^2\implies+-2=y+3`

Given that there will be 2 points on the y-axis that the equation will map through. So, taking the above equation and making 2 equations such for the two points, we get
`+2=y_1+3` and `-2=y_2+3`

Given that we have to find the distance between the 2 points, and hence we have to do `y_1-y_2`.
So, from the above equation `2-3=y_1` and `-2-3=y_2`
Doing as such, we get `4=y_1-y_2`

The 2 points `y_1` and `y_2` are the very intercepts, and hence the distance will be the difference of their values, which we have got is `4`.

Graph for proof.
graph{x+8=2(y+3)^2 [-10.16, 9.84, -7.58, 2.42]}