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Mathematics 88 Online
OpenStudy (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.

OpenStudy (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

OpenStudy (anonymous):

thank you @Nnesha :D

Nnesha (nnesha):

x=-2 , x=3 set it equal to zero

OpenStudy (anonymous):

okay... then do I plug in something?

Nnesha (nnesha):

then you would apply foil method to get the standard form

OpenStudy (anonymous):

foil method? o.o//

Nnesha (nnesha):

(a+b)(c+d) multiply ? FOIL familiar with it ?

OpenStudy (anonymous):

-2^2 +3+-4=

OpenStudy (anonymous):

??? cross multiply, sorry I don't really remember...

OpenStudy (anonymous):

...

OpenStudy (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

OpenStudy (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

OpenStudy (anonymous):

a= - y/2x(x-3)

OpenStudy (anonymous):

a=−y2x(x−3)

OpenStudy (anonymous):

okay :/

OpenStudy (anonymous):

@Robert136 please help?

Nnesha (nnesha):

replace y with -4

Nnesha (nnesha):

read the statement of the original question

OpenStudy (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

OpenStudy (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

OpenStudy (anonymous):

y= 1/2 (x --2) (x--3)?

Nnesha (nnesha):

a =2/3 not 1/2

OpenStudy (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)

OpenStudy (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 = ?

OpenStudy (anonymous):

-2? 3

Nnesha (nnesha):

after that you wll with part a

Nnesha (nnesha):

negative times negative = positive

Nnesha (nnesha):

so -1 times -2 = ?

OpenStudy (anonymous):

2

Nnesha (nnesha):

yes so -1 times 3 = ?

OpenStudy (anonymous):

-3?

Nnesha (nnesha):

right

OpenStudy (anonymous):

okay xD

OpenStudy (anonymous):

2, -3

Nnesha (nnesha):

\[\huge\rm y=\frac{ 2 }{ 3 }(x+2))(x-3)\]

OpenStudy (anonymous):

y=2/3(x+2)(x-3)

OpenStudy (anonymous):

lol

Nnesha (nnesha):

yep!

OpenStudy (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

OpenStudy (anonymous):

sorry I had to go the restroom...

Nnesha (nnesha):

are you there ?

Nnesha (nnesha):

tag me when u're ready okay :)

OpenStudy (anonymous):

hey

OpenStudy (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)

OpenStudy (anonymous):

okay hold up

OpenStudy (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

OpenStudy (anonymous):

okay

Nnesha (nnesha):

let me know what u get

OpenStudy (anonymous):

I get nothing :/

Nnesha (nnesha):

what do you mean ?

OpenStudy (anonymous):

I get 0

OpenStudy (anonymous):

also sorry I had to leave the computer :/

Nnesha (nnesha):

how did you get 0 ?? 2/3(x-3)(x+2) ??

OpenStudy (anonymous):

multiplied....

OpenStudy (anonymous):

2x^2/3 - 2x/3 -4???

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