Okay I have a math exam (mock) tomorrow.. and I was doing revision when I had this question, I've never really got how to do this type of question (how to put the 1/2 into the brackets)...
My guess is:
y= ½ (x-4) (x+2)
y= ½x²-2x-2 Even if this is right once I get to here I don't know how to put it back in brackets without endless guess & checking...
When I put the equation into my graphics calculator it has the (x-4)(x-2) part with the x-intercepts at 4 and -2 but- Yay! I just just figured out how it works for GRAPHS. My caclulator has got the vertex at -4.5, half of the 3².... duh...
Hmm.. so I know how the graph works but I really don't get the equation, so my question still stands, how do you do an equation like this?
y= ½ (x-4) (x+2)
2007-08-13 22:57:08 · 5 answers · asked by Holly from NZ 2 in Science & Mathematics ➔ Mathematics
Re Awesomely Lame, whoa well then year 8 I'm year 11 so you'd have to be damn a smart year 8 if you did know that! The wonderful world of quadratics... haha..
2007-08-13 23:21:54 · update #1
y = (1/2) (x - 4) (x + 2)
Cuts y axis when x = 0
ie at point A(0 , - 4)
Cuts x axis when y = 0
ie when (x - 4) (x - 2) = 0
x = 4 , x = - 2
points B(4 , 0) , C(- 2 , 0)
f(x) = (1/2) (x² - 2x - 8)
f(x) = (1/2)x² - x - 4
f `(x) = x - 1 = 0 for turning point
x = 1 for turning point.
D(1 , - 9/2) is a turning point
Join A , B , C and D to obtain graph.
2007-08-17 22:56:02 · answer #1 · answered by Como 7 · 0⤊ 0⤋
1/2(x - 4)(x + 2) = 1/2(x^2 - 2x - 4) = (x^2)/2 - x - 2
to put it back to the brackets, take the common factor of 1/2 out
1/2 ( x^2 - 2x - 4)
then factorise the backet
1/2 ( x - 4)(x + 2)
the x intercepts should stay the same, since 1/2(x - 4)(x + 2) = 0
is also the same as (x - 4)(x + 2) = 0 is u just times both sides by 2
Compare (x^2)/2 - x - 2 with x^2 - 2x - 4 , u should notice that the y intercept is halved. Simply put it this way. For every x values, the y values are halved (compared to (x - 4)(x + 2)) because the equation can also be written as 2y = (x - 4)(x + 2).
The x intercepts stays the same because half of 0 is still 0.
2007-08-13 23:12:58 · answer #2 · answered by Anonymous · 1⤊ 0⤋
no... see what u do is multiply the brackets first and then the .5 into the brackets... sooo for the one u did
.5*(x²-2x-8)
then it simplifies to .5x²-x-4... litterally half of what it used to be....
for the vertex i have no clue how u got 4.5 .... its rather simple -b/2a where ax²+bx+c is ur equasion... but maybe u got mixed up with somin else
the x inter are easy too ... for each section with parenthesis make it equal 0... sooo if it says (x-3) then one of em is 3 or (x+1) the its -1
lastly how you "do" somin like that ... its a basic u ... if u have the zero's which looks like u understand ... and the vertex... which i explained... then u have 3 dots ... connect them...
if ur teach gets mean and gives u only one dot or no dots just ...main thing is to have points arround the vertex tho so u can judge how fast its gonna go up so choose one x up and one down and just enter them into ur equasion
2007-08-13 23:16:46 · answer #3 · answered by R 2 the T 2 · 1⤊ 0⤋
It is a parabol with a minimum at x=-1.5 its axis corresponds to x=-b/2a= -12/8 =-1.5 y = -10 the roots are x=0 and x=-3 . for these values y=0 see my link
2016-05-17 08:58:11 · answer #4 · answered by ? 3 · 0⤊ 0⤋
we are doing that in maths right now... but i have only learnt to do quadratic ones by turning them into factors (brackets) not expanding them. I don't get it because the way you did it seemed right, but then it seems (to me... only yr 8 here lol) that it is impossible because you can't factorise the answer you came up with... ask your teacher?
2007-08-13 23:06:23 · answer #5 · answered by awesomely_lame 3 · 0⤊ 0⤋