Question

In which direction does the parabola y= 1/2 (x–1)2–7 open?

Answer 1

Opens up

Explanation:

Generally parabolas open up/down when you have an equation like this:
`y=a(x-h)^2+k`

and when parabolas open to the left/right you have an equation as such:
`x=a(y-k)^2+h`

Btw the k and h being in different places is not a typo.

Anyways the reason for why
`y=a(x-h)^2+k` opens up, is for each x-value, you're only going to have one y-value. This is a parabola, so it's not a one-to-one function (where each output is unique), but it's still a function regardless.

But when you have this equation:
`x=a(y-k)^2+h`, there are two possible y-values, that will output the same x-value. Or in other words, each x-value (except the vertex) will have two outputs. So it's going to open sideways.

Anyways in the equation you provided we have the form:
`y=a(x-h)^2+k` so the parabola is opening up or down. Now the only thing that really determines this is the "a" term"

when a > 0 the parabola opens up

when a < 0 the parabola opens down

Note: This only applies when the parabola opens up/down

Since we have a positive "a" term, the parabola opens up. This makes sense, since the base of the exponent:
`(x-h)` is going to be growing faster than the constant:
`-7`, so as
`x\implies \infty` the function will be increasing. It is of course decreasing as
`x\implies1`, since at the vertex that's the minimum. The reason for this is because:
`f(h-x)=f(h+x)` due to the symmetry of a parabola. So in our specific case:
`f(1+x)=f(1-x)`, this means that despite f(-100) having a lower x-value than f(50), it actually has a greater y-value, and has the same y-value as: f(102), since f(1-101) = f(1+101); meaning f(-100) = f(102). So as x goes towards zero, it's actually decreasing in this case. But after it passes the vertex, it will increase and go towards positive infinity.