## anonymous one year ago Pre Calc Help

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1. anonymous

2. KaiserTheSlayer

do want to know how or the answer??

3. anonymous

4. KaiserTheSlayer

ok

5. anonymous

i know it is steepness but idk the exact rule

6. anonymous

can anyone help?

7. KaiserTheSlayer

yes but it would take a little

8. anonymous

thats fine

9. anonymous

I just need someone to explain how i know A based off looking at graph i know its negative but how do i know if its 1,2, etc

10. KaiserTheSlayer

alright, so f(x)=-(x-1)^2+4, does it make sense why

11. anonymous

can you give it in standard form

12. anonymous

ax^2 + bx + c

13. anonymous

and no im not sure how you got that

14. KaiserTheSlayer

okay so since we know the vertext, we know how much to shift it, up four and to the right 1, so thats how i got that

15. KaiserTheSlayer

to get the form you want, factor and simplfy it

16. anonymous

wait so if that was a steeper parabola what would you change

17. KaiserTheSlayer

the a and for the answer it is -1

18. anonymous

where is a in the formula you gave me?

19. Nnesha

alright you can find vertex point by looking at the graph right then substitute into this equation $\huge\rm y = a(x-h)^2 +k$ vertex form of quadratic equation after that pick x and y values plug in solve for a :-)

20. KaiserTheSlayer

-x^2+2x+3

21. Nnesha

(x,y) (0,3)

22. anonymous

when y=0,x=-1 and 2

23. Nnesha

and by looking at the shape of graph u can easily find if a is negative or positive

24. anonymous

correction no,x=-1 and x=3

25. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @MathMan77 where is a in the formula you gave me? $$\color{blue}{\text{End of Quote}}$$ $$\color{blue}{\text{Originally Posted by}}$$ @KaiserTheSlayer alright, so f(x)=-(x-1)^2+4, does it make sense why $$\color{blue}{\text{End of Quote}}$$ $y=-(x-1)^2+4$ --- >$\huge\rm y = -\color{red}{a}(x-h)^2 +k$ invisble one

26. anonymous

ah thank you!

27. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @Nnesha alright you can find vertex point by looking at the graph right then substitute into this equation $\huge\rm y = a(x-h)^2 +k$ vertex form of quadratic equation after that pick x and y values plug in solve for a :-) $$\color{blue}{\text{End of Quote}}$$ vertex point ( 1,4) (x,y) (0,3) $\large\rm 3=a(0-1)^2 +4$ like this if u solve for a u will get the answer

28. anonymous

y=-(x+1)(x-3)=-(x^2-2x-3)=-(x^2-2x)+3=-(x^2-2x+1-1)+3=-(x^2-2x+1)+1+3 =-(x-1)^2+4