anonymous
  • anonymous
Medal and explain as well please. :) ♥ Question: A parabola has intercepts of x = -2, x = 3, and y = -4. Answer these a. Write the intercept form of the parabola. b. State the direction of the parabola. Explain. c. Write in ax2 + bx + c form. d. What is the axis of symmetry? e. What is the vertex? f. Graph the parabola. Label axis, vertex, and intercepts.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@Nnesha please help?
Nnesha
  • Nnesha
what is the intercept form ? ax^2+bx+c = standard form y=a(x-h)^2+k vertex form
anonymous
  • anonymous
thank you @Nnesha :D

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More answers

Nnesha
  • Nnesha
x=-2 , x=3 set it equal to zero
anonymous
  • anonymous
okay... then do I plug in something?
Nnesha
  • Nnesha
then you would apply foil method to get the standard form
anonymous
  • anonymous
foil method? o.o//
Nnesha
  • Nnesha
(a+b)(c+d) multiply ? FOIL familiar with it ?
anonymous
  • anonymous
-2^2 +3+-4=
anonymous
  • anonymous
??? cross multiply, sorry I don't really remember...
anonymous
  • anonymous
...
anonymous
  • anonymous
y=(x^2+x^3)^4 ???
Nnesha
  • Nnesha
this is the intercept form of parabola \[\huge\rm y=a(x-p)(x-q)\] where p and q are the x-intercepts y= y-intercept
Nnesha
  • Nnesha
a)^^
Nnesha
  • Nnesha
given x-intercepts are -2 and 3 so substitute p and q for -2 and 3
Nnesha
  • Nnesha
\[\huge\rm y=a(x-(-2))(x-3)\] y intercept is (0 , -4) { y-intercept is a point where graph of the equation intersect y-axis when x=0 ) so substitute y for -4 and 0 for x solve for a
anonymous
  • anonymous
I never really learned this, so don't be mad... Anyways okay
Nnesha
  • Nnesha
i've never heard abt intercept form this is first time i'm doing this
Nnesha
  • Nnesha
feel free to ask question if you don't get anything :)
Nnesha
  • Nnesha
http://www.virtualnerd.com/algebra-2/quadratics/transforming-functions/intercept-form/intercept-form-from-graph i watched this video watch this maybe you will understand like i said today is the first time i'm solving intercept form already familiar with vertex , standard form but not the intercept :D
anonymous
  • anonymous
a= - y/2x(x-3)
anonymous
  • anonymous
a=−y2x(x−3)
anonymous
  • anonymous
okay :/
anonymous
  • anonymous
@Robert136 please help?
Nnesha
  • Nnesha
replace y with -4
Nnesha
  • Nnesha
read the statement of the original question
anonymous
  • anonymous
a=-4/2x(x-3)
Nnesha
  • Nnesha
\[-4=a(x-(-2))(x-3)\] when y =-4 x would be 0 bec when line intersect at y-axis x=0 so replace x with 0
Nnesha
  • Nnesha
\[\huge\rm -4=a(\color{reD}{0}-(-2))(\color{reD}{0}-3)\] like this now solve for a
anonymous
  • anonymous
a=2/3
Nnesha
  • Nnesha
yes right so intercept for would be ? \[\huge\rm y=a(x-p)(x-q)\] replace a with 2/3 p =-2 q=-3
anonymous
  • anonymous
y= 1/2 (x --2) (x--3)?
Nnesha
  • Nnesha
a =2/3 not 1/2
anonymous
  • anonymous
opps xD i don't know where that came from xD
Nnesha
  • Nnesha
\[\huge\rm y=\frac{ 2 }{ 3 }(x\color{reD}{-(-2)})(x-3)\] distribute -(-2)
anonymous
  • anonymous
y=2/3(x--2)(x--3)
Nnesha
  • Nnesha
hmm q isn't -3
Nnesha
  • Nnesha
\[\huge\rm y=\frac{ 2 }{ 3 }(x\color{reD}{-(-2)})(x-(+3))\] distribute -(-2)
Nnesha
  • Nnesha
-1 times -2 = ? - times +3 = ?
anonymous
  • anonymous
-2? 3
Nnesha
  • Nnesha
after that you wll with part a
Nnesha
  • Nnesha
negative times negative = positive
Nnesha
  • Nnesha
so -1 times -2 = ?
anonymous
  • anonymous
2
Nnesha
  • Nnesha
yes so -1 times 3 = ?
anonymous
  • anonymous
-3?
Nnesha
  • Nnesha
right
anonymous
  • anonymous
okay xD
anonymous
  • anonymous
2, -3
Nnesha
  • Nnesha
\[\huge\rm y=\frac{ 2 }{ 3 }(x+2))(x-3)\]
anonymous
  • anonymous
y=2/3(x+2)(x-3)
anonymous
  • anonymous
lol
Nnesha
  • Nnesha
yep!
anonymous
  • anonymous
xD ♥
Nnesha
  • Nnesha
now multiply 2/3(x+2)(x-3) and i've to go now sorry its urgent :(
Nnesha
  • Nnesha
for part b) if a is positive then you will get`u` shape and if a is negative then you will hve `n` shape
Nnesha
  • Nnesha
cya
anonymous
  • anonymous
sorry I had to go the restroom...
Nnesha
  • Nnesha
are you there ?
Nnesha
  • Nnesha
tag me when u're ready okay :)
anonymous
  • anonymous
hey
anonymous
  • anonymous
still here sups, aha
Nnesha
  • Nnesha
okay great ready !?
Nnesha
  • Nnesha
`Write in ax2 + bx + c form. ` multiply 2/2(x-3)(x+2)
anonymous
  • anonymous
okay hold up
anonymous
  • anonymous
x^2-x-6?
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @Nnesha `Write in ax2 + bx + c form. ` multiply 2/2(x-3)(x+2) \(\color{blue}{\text{End of Quote}}\) 2/3**
Nnesha
  • Nnesha
(x+2)(x-3)=x^2-x-6 now multiply it by 2/3
anonymous
  • anonymous
okay
Nnesha
  • Nnesha
let me know what u get
anonymous
  • anonymous
I get nothing :/
Nnesha
  • Nnesha
what do you mean ?
anonymous
  • anonymous
I get 0
anonymous
  • anonymous
also sorry I had to leave the computer :/
Nnesha
  • Nnesha
how did you get 0 ?? 2/3(x-3)(x+2) ??
anonymous
  • anonymous
multiplied....
anonymous
  • anonymous
2x^2/3 - 2x/3 -4???

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