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.

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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
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@Nnesha please help?
what is the intercept form ? ax^2+bx+c = standard form y=a(x-h)^2+k vertex form
thank you @Nnesha :D

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Other answers:

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

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