Pre Calc Help

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- anonymous

Pre Calc Help

- jamiebookeater

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- anonymous

##### 1 Attachment

- KaiserTheSlayer

do want to know how or the answer??

- anonymous

both please

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## More answers

- KaiserTheSlayer

ok

- anonymous

i know it is steepness
but idk the exact rule

- anonymous

can anyone help?

- KaiserTheSlayer

yes but it would take a little

- anonymous

thats fine

- 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

- KaiserTheSlayer

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

- anonymous

can you give it in standard form

- anonymous

ax^2 + bx + c

- anonymous

and no im not sure how you got that

- 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

- KaiserTheSlayer

to get the form you want, factor and simplfy it

- anonymous

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

- KaiserTheSlayer

the a and for the answer it is -1

- anonymous

where is a in the formula you gave me?

- 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
:-)

- KaiserTheSlayer

-x^2+2x+3

- Nnesha

(x,y) (0,3)

- anonymous

when y=0,x=-1 and 2

- Nnesha

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

- anonymous

correction
no,x=-1 and x=3

- 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

- anonymous

ah thank you!

- 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

- 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

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