Answer:
Opens up
Step-by-step explanation:
Generally parabolas open up/down when you have an equation like this: [tex]y=a(x-h)^2+k[/tex]
and when parabolas open to the left/right you have an equation as such: [tex]x=a(y-k)^2+h[/tex]
Btw the k and h being in different places is not a typo.
Anyways the reason for why [tex]y=a(x-h)^2+k[/tex] 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: [tex]x=a(y-k)^2+h[/tex], 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: [tex]y=a(x-h)^2+k[/tex] 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: [tex](x-h)[/tex] is going to be growing faster than the constant: [tex]-7[/tex], so as [tex]x\implies \infty[/tex] the function will be increasing. It is of course decreasing as [tex]x\implies1[/tex], since at the vertex that's the minimum. The reason for this is because: [tex]f(h-x)=f(h+x)[/tex] due to the symmetry of a parabola. So in our specific case: [tex]f(1+x)=f(1-x)[/tex], 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.
