## iwanttogotostanford one year ago ACT PREP Q/ PLEASE HELPPP: ACT PREP Q/ PLEASE HELP: The functions f(x) = –(x + 4)^2 + 2 and g(x) = (x − 2)^2 − 2 have been rewritten using the completing-the-square method. Is the vertex for each function a minimum or a maximum? Explain your reasoning for each function.

1. iwanttogotostanford

@zepdrix

2. zepdrix

$$\large\rm y=x^2$$ is a parabola opening upward.|dw:1443654118776:dw|

3. zepdrix

|dw:1443654152110:dw|So the vertex is a minimum. It is lower than any other point on the curve.

4. zepdrix

$$\large\rm y=-x^2$$ is a parabola opening downward.|dw:1443654195359:dw|

5. zepdrix

|dw:1443654246434:dw|And we can clearly see that the vertex is the highest point on the curve, the maximum.

6. zepdrix

Hmmmm so what do you think? :d how does that apply to f(x) and g(x)

7. iwanttogotostanford

im not sure, i don't know how to do this at all- i am studying for the ACT and was so clueless on this one!

8. iwanttogotostanford

I have not learned this yet and this was one of the harder questions for the ACT practice

9. iwanttogotostanford

@zepdrix

10. zepdrix

well use my examples :O and try to guess at least

11. iwanttogotostanford

So I was hoping for a step by step so I could use it for learning! :-)

12. zepdrix

u -_-

13. iwanttogotostanford

What??! I have an ACT prep book right in front of me and I am clueless! Its not for a grade

14. zepdrix

no, im saying, give ME a guess. lol

15. zepdrix

do you understand how to find the vertex when a function is given in this form? :)

16. iwanttogotostanford

ok, well I think the vertex is a minimum and NOPE

17. iwanttogotostanford

I am learning algebra 2 right noe but I have the ACT this october

18. zepdrix

well unfortunately, this is two questions, not just one. It's asking about two separate functions, f(x) and g(x).

19. iwanttogotostanford

i know! I need helpp

20. iwanttogotostanford

21. zepdrix

the vertex form of a parabola looks like this:$\large\rm y=(x-\color{#3366CF}{h})^2+\color{#3366CF}{k}$Where the vertex is located at $$\large\rm (\color{#3366CF }{h},~\color{#3366CF }{k})$$. But what is interesting is.... the location of the vertex actually has no impact on whether it is a maximum or minimum!

22. iwanttogotostanford

huh thats interesting!

23. zepdrix

So for our function:$\large\rm f(x)=-(x-\color{#3366CF}{(-4)})^2+\color{#3366CF}{2}$We can see that our vertex is located at $$\large\rm (\color{#3366CF }{-4},~\color{#3366CF }{2})$$. But this doesn't matter so much. All we really care about, is whether or not there is a negative sign in front of the squared term.

24. iwanttogotostanford

oh ok

25. zepdrix

Negative sign means the parabola opens down, means vertex is a maximum. positive sign means the parabola opens up, meaning the vertex is a minimum. For f(x), the squared term has a negative in front. So our vertex corresponds to a maximum. How bout g(x)? Any ideas? :o

26. iwanttogotostanford

@zepdrix minimum right? for g(x)

27. zepdrix

$\large\rm g(x)=(x-2)^2-2$Mmmm ok good! There are a lot of negatives floating around in g, so it's easy to get confused, but what's important is that the square is positive,$\large\rm g(x)=+(x-2)^2-2$So the vertex of g(x) corresponds to a minimum. yay team \c:/

28. zepdrix

XD

29. iwanttogotostanford

@zepdrix

30. iwanttogotostanford

@zepdrix

Find more explanations on OpenStudy