Question

Okay so a graph has x-intercepts -3, -0.5 1 and 2, and it asks you to find the equation of the function, which passes through the point (-1,-6). So I wrote the parent function f(x)=a(x+3)(x+0.5)(x-1)(x-2). However, when i substituted (-1,-6) I got a different value for a as the textbook did, and it's because they used (2x+1) as a factor instead of (x+0.5). So does that mean that I you can't use decimals in the factor?

Answer 1

since those are your zeros, multiply them up

Answer 2

given intercepts at: a,b,c,d, then:
(x-a)(x-b)(x-c)(x-d)=0 will give you your basic equation

Answer 3

then determine that value of that basic for x=-1 and see what you have to scale by to get it to -6

Answer 4

decimals are fine for factors; but can be written in other ways

Answer 5

the issue is in the scalar; most likely its a 2 to account for that

Answer 6

what do you mean by scalar?

Answer 7

its that "a" in front

you have a basic equation that can be scaled up or down to fine tune it and keep the same intercepts

Answer 8

x^2 -1 has the same intercepts as 2(x^2-1) but the graph hits a new set of points everywhere else

Answer 9

http://www.wolframalpha.com/input/?i=%28x%2B3%29%28x%2B0.5%29%28x-1%29%28x-2%29%2C+x%3D-1

the wolf say the basic equation is good; x=-1 is approx -6

Answer 10

i see, with the 2x+1 part youve actually doubled it so you have to divide it back out for the "a"

Answer 11

yeah that's the problem :(

Answer 12

at any rate, your setup was fine.

Answer 13

i don't know whether i should use x+0.5 or 2x+1

Answer 14

2(x+.5) = (2x+1) so its really just a matter of what the author of your text has decided to go with

Answer 15

2x+1 = 0 when x=-.5

x+.5 = 0 when x = -.5

Answer 16

their choice to double it to begin with accounts for a difference in the scalars you choose for "a"

Answer 17

but after everything is expanded out, it should be the same

Answer 18

yours
\[x^4+0.5 x^3-7. x^2+2.5 x+3\]
theirs
\[x^4+x^3/2-7 x^2+(5 x)/2+3\]

Answer 19

oh kk thanks!