anonymous
  • anonymous
Pre Calc Help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
KaiserTheSlayer
  • KaiserTheSlayer
do want to know how or the answer??
anonymous
  • anonymous
both please

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KaiserTheSlayer
  • KaiserTheSlayer
ok
anonymous
  • anonymous
i know it is steepness but idk the exact rule
anonymous
  • anonymous
can anyone help?
KaiserTheSlayer
  • KaiserTheSlayer
yes but it would take a little
anonymous
  • anonymous
thats fine
anonymous
  • 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
  • KaiserTheSlayer
alright, so f(x)=-(x-1)^2+4, does it make sense why
anonymous
  • anonymous
can you give it in standard form
anonymous
  • anonymous
ax^2 + bx + c
anonymous
  • anonymous
and no im not sure how you got that
KaiserTheSlayer
  • 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
  • KaiserTheSlayer
to get the form you want, factor and simplfy it
anonymous
  • anonymous
wait so if that was a steeper parabola what would you change
KaiserTheSlayer
  • KaiserTheSlayer
the a and for the answer it is -1
anonymous
  • anonymous
where is a in the formula you gave me?
Nnesha
  • 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
  • KaiserTheSlayer
-x^2+2x+3
Nnesha
  • Nnesha
(x,y) (0,3)
anonymous
  • anonymous
when y=0,x=-1 and 2
Nnesha
  • Nnesha
and by looking at the shape of graph u can easily find if a is negative or positive
anonymous
  • anonymous
correction no,x=-1 and x=3
Nnesha
  • 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
  • anonymous
ah thank you!
Nnesha
  • 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
  • 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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