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Wondermath
 4 years ago
Okay so a graph has xintercepts 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)(x1)(x2). 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?
Wondermath
 4 years ago
Okay so a graph has xintercepts 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)(x1)(x2). 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?

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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1since those are your zeros, multiply them up

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1given intercepts at: a,b,c,d, then: (xa)(xb)(xc)(xd)=0 will give you your basic equation

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1then determine that value of that basic for x=1 and see what you have to scale by to get it to 6

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1decimals are fine for factors; but can be written in other ways

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the issue is in the scalar; most likely its a 2 to account for that

Wondermath
 4 years ago
Best ResponseYou've already chosen the best response.0what do you mean by scalar?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1its 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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1x^2 1 has the same intercepts as 2(x^21) but the graph hits a new set of points everywhere else

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=%28x%2B3%29%28x%2B0.5%29%28x1%29%28x2%29%2C+x%3D1 the wolf say the basic equation is good; x=1 is approx 6

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i see, with the 2x+1 part youve actually doubled it so you have to divide it back out for the "a"

Wondermath
 4 years ago
Best ResponseYou've already chosen the best response.0yeah that's the problem :(

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1at any rate, your setup was fine.

Wondermath
 4 years ago
Best ResponseYou've already chosen the best response.0i don't know whether i should use x+0.5 or 2x+1

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.12(x+.5) = (2x+1) so its really just a matter of what the author of your text has decided to go with

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.12x+1 = 0 when x=.5 x+.5 = 0 when x = .5

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1their choice to double it to begin with accounts for a difference in the scalars you choose for "a"

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1but after everything is expanded out, it should be the same

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1yours \[x^4+0.5 x^37. x^2+2.5 x+3\] theirs \[x^4+x^3/27 x^2+(5 x)/2+3\]
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