# Find pattern, equation, then graph...

• September 13th 2007, 12:44 PM
trancefanatic
Find pattern, equation, then graph...
Directions: Find the pattern in each problem. Find the equation. Graph the line on a coordinate plane.

********
i'm in a class that mixes algebra 2 w/pre-cal and i have absolutely no idea on how to proceed on finding the equation. on #2 i noticed a +2 pattern on the second row. thanks in advance
• September 13th 2007, 12:52 PM
Jhevon
Quote:

Originally Posted by trancefanatic
Directions: Find the pattern in each problem. Find the equation. Graph the line on a coordinate plane.

********
i'm in a class that mixes algebra 2 w/pre-cal and i have absolutely no idea on how to proceed on finding the equation. on #2 i noticed a +2 pattern on the second row. thanks in advance

here's a hint for the first, the rest seem to be similar.

note that since for each new term we are adding or subtracting a constant number, we can describe the pattern by an arithmetic sequence.

recall that an arithmetic sequence is one of the form: $a_n = a_1 + (n - 1)d$

where $a_n$ is the nth term, $a_1$ is the first term, $n$ is the current number of the term, and $d$ is the common difference (that is, the constant we add to one term to get the next term).

fill in the corresponding value in the formula and simplify. the result will be in the form of the equation of a line

an alternative is to use the point-slope form of the equation of a line to find the answer.

the point slope form is: $y - y_1 = m(x - x_1)$, where $m$ is the slope, which we can find from any two points, and $(x_1,y_1)$ is a point the line passes through (in this case, of course, n is the input value, so n would replace x in the formulas)

got it?
• September 13th 2007, 02:02 PM
trancefanatic
for #2, i got 2n-7. but when i tried to create an equation, i got m=-8 and on the second attempt with another pair of coordinates, i got m=2. then i thought it'd be something like y=2x-7... i think i have an idea.

• September 13th 2007, 02:25 PM
Jhevon
Quote:

Originally Posted by trancefanatic
for #2, i got 2n-7. but when i tried to create an equation, i got m=-8 and on the second attempt with another pair of coordinates, i got m=2. then i thought it'd be something like y=2x-7... i think i have an idea.

wait, i'll see what i get

Quote:

i doubt it. i'd try to fit it to a cubic. do you know how? Hint: we can use simultaneous equations

EDIT: ok, i got (i wrote them as the lines you will plot, so i replaced n with x and a_n with y):

#1: y = 2 - 3x

#2: y = 2x - 7

#3: y = 2 - 2x
• September 13th 2007, 02:29 PM
trancefanatic
i hated simultaneous equations and never did understand it... you can graph those??
• September 13th 2007, 02:31 PM
Jhevon
Quote:

Originally Posted by trancefanatic
i hated simultaneous equations and never did understand it... you can graph those??

yes, they are straight lines. they are relatively easy to graph, especially in comparison to a cubic, which is what i believe #4 is.

to graph the lines, find the x- and y-intercepts, plot them on the axis, and draw a straight line through them, that's it. you do know how to find the intercepts, right?
• September 13th 2007, 02:34 PM
trancefanatic
Quote:

Originally Posted by Jhevon

EDIT: ok, i got (i wrote them as the lines you will plot, so i replaced n with x and a_n with y):

#1: y = 2 - 3x

#2: y = 2x - 7

#3: y = 2 - 2x

i got
y=2x-7
y=-2x+2
y=-3x+2

which i think are the same as yours. on #4 i tried to see if there were any patterns and ended up factoring 18. but that's pretty much it...
• September 13th 2007, 02:35 PM
trancefanatic
Quote:

Originally Posted by Jhevon
yes, they are straight lines. they are relatively easy to graph, especially in comparison to a cubic, which is what i believe #4 is.

to graph the lines, find the x- and y-intercepts, plot them on the axis, and draw a straight line through them, that's it. you do know how to find the intercepts, right?

yes, i realized that now that i thought about what simultaneous equations looks like.
• September 13th 2007, 02:39 PM
Jhevon
Quote:

Originally Posted by trancefanatic
yes, i realized that now that i thought about what simultaneous equations looks like.

a hint for #4:

Let the cubic be of the form: $y = ax^3 + bx^2 + cx + d$

Note that: $f(1) = 6$, $f(2) = 2$, $f(3) = 40$, and $f(4) = 60$

do you think you can take it from here? to set up the simultaneous equations and solve them?
• September 13th 2007, 02:52 PM
trancefanatic
Quote:

Originally Posted by Jhevon
a hint for #4:

Let the cubic be of the form: $y = ax^3 + bx^2 + cx + d$

Note that: $f(1) = 6$, $f(2) = 2$, $f(3) = 40$, and $f(4) = 60$

do you think you can take it from here? to set up the simultaneous equations and solve them?

i'm definitely lost...
• September 13th 2007, 03:01 PM
Jhevon
Quote:

Originally Posted by trancefanatic
i'm definitely lost...

well, $f(x) = y$, so we have $f(x) = ax^3 + bx^2 + cx + d$

" $f(1) = 6$" means, when $x = 1, y= 6$

so since $f(1) = 6$, we have:

$6 = a(1)^3 + b(1)^2 + c(1) + d = a + b + c + d$

since $f(2) = 2$, we have:

$2 = a(2)^3 + b(2)^2 + c(2) + d = 8a + 4b + 2c + d$

since $f(3) = 40$, we have:

$40 = a(3)^3 + b(3)^2 + c(3) + d = 27a + 9b + 3c + d$

since $f(4) = 60$, we have:

$60 = a(4)^3 + b(4)^2 + c(4) + d = 64a + 16b + 4c + d$

Thus, to find the coefficients in our cubic, we need to solve the system:

$6 = a + b + c + d$ ......................(1)

$2 = 8a + 4b + 2c + d$ .................(2)

$40 = 27a + 9b + 3c + d$ ..............(3)

$60 = 64a + 16b + 4c + d$ ............(4)

Hopefully you can do that
• September 13th 2007, 03:11 PM
trancefanatic
OH ok now i get it. you didn't have to do them all, but i really appreciate it! thanks. i'll probably redo them just for practice and comprehension check. :)