Question

Pre Calc Help

Answer 1

do want to know how or the answer??

Answer 2

both please

Answer 3

ok

Answer 4

i know it is steepness
but idk the exact rule

Answer 5

can anyone help?

Answer 6

yes but it would take a little

Answer 7

thats fine

Answer 8

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

Answer 9

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

Answer 10

can you give it in standard form

Answer 11

ax^2 + bx + c

Answer 12

and no im not sure how you got that

Answer 13

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

Answer 14

to get the form you want, factor and simplfy it

Answer 15

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

Answer 16

the a and for the answer it is -1

Answer 17

where is a in the formula you gave me?

Answer 18

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
:-)

Answer 19

-x^2+2x+3

Answer 20

(x,y) (0,3)

Answer 21

when y=0,x=-1 and 2

Answer 22

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

Answer 23

correction
no,x=-1 and x=3

Answer 24

\(\color{blue}{\text{Originally Posted by}}\) @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

Answer 25

ah thank you!

Answer 26

\(\color{blue}{\text{Originally Posted by}}\) @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

Answer 27

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