Hi Mike, Can you please explain Question 11, Test 6, section 3 ? I know the parabola opens downward, but I’m confused after that. Thanks.

You’ve already got the first part—when you add a negative to the front of a parabola you’ll flip it vertically. Since this parabola begins facing up, the negative flips it so it faces down.

From there, I think it’s helpful to think about this in terms of how functions shift. For some function f(x) that’s graphed on the xy-plane:

  • f(x)+1 ⇒ (graph moves UP one)
  • f(x)–1 ⇒ (graph moves DOWN one)
  • f(x+1) ⇒ (graph moves LEFT one)
  • f(x–1) ⇒ (graph moves RIGHT one)

That applies here. You’re basically taking the ax^2 function and applying a few shifts (and a flip). Check it out–if you start by assuming a function p(x)=ax^2, you can build the function in this question by shifting that original function:


-p(x-b)+c flips the function p(x) upside-down, then shifts it b units to the right and c units up.

If you hate everything I’ve just said, I have good news! You can also just memorize the vertex form of a parabola (which we basically just derived).

If you have a parabola in the form f(x)=a(x-h)^2+k, then you know it has its vertex at (h, k) and that the sign of a tells you whether the parabola opens up or down. This question basically gives you the vertex form, only it uses b and c instead of h and k. Recognize that and you know right away that the parabola’s vertex is at (b, c).

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