unit+6

For each of the groups below, identify the graph that does not belong and state your reasoning why that graph does not belong in your online journal.

graph 3: Does not belong because the end behavior wrong direction

group 2: doe not belong because of the linear function and the diffrent end behavior

group 4: graph three fits because its a linear function

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__** 6.2: **__
== == Summarize the last two days of class in your online journal. We have discussed different methods for graphing polynomial functions in intercept form. In detail, explain the graphing method to a student who has missed the last two days.

Whenever graphing polynomials one needs to find the x-intercepts by having the parenthesis equal to 0. Finding the y-intercept is the next process where one just plugs 0 in for the variable x. The __degree__ and the leading coefficient are the last two things. It tells one the end behavior allowing the completion of the graph.

=__ 6.6 __=

q(x) and u(x) match graph 6 because:
1)q(x) is in an intercept form i have to add all the degree. 2) u(x) is in standard form i choose the first degree and i take the highest degree 3) both have negative leading coefficient with an even degree. 4) their end behavior are bottom left bottom right. 5) since there were two graphs who were going down i found each y intercept i plug 0 for x . q(x) and u(x) equal both 20.

k(x) and a(x) match graph 2 because:
1)k(x) is an intercept form i have to add all the degree 2)a(x) is in standard form i choose the first degree i take the highest exponent 3) both have positive leading coefficent and odd degree. 4) their end behavior are bottom left top right 5) in graph 2 top right was going up to cross the y intercept at 15.

n(x) and i(x) match graph 4 because:
1) n(x) is in an intercept form i have to add all the degree 2) i(x) is in standard form i choose the first degree i take the highest exponent 3)both have postive leading coefficient and even degree 4) their end behavior are top left top right. 5) i plug 0 for x to find positive 6 for y intercept.

p(x) and l(x) match graph 5 because:
1)p(x) is in an intecept form i have to add all the degree 2)l(x) is in a standard form i choose the first degree i take the highest exponent 3) both have negative leAding coefficient and even degree 4) their end behavior are bottom left bottom right. 5) i plug 0 for x to find -6 for y intercept on the graph

m(x) and e(x) match graph 3 because
1)m(x) is in an intercept form i have to add all the degree 2) e(x) is in a standard form i choose the first degree i take the highest exponent 3) both have negative leading odd degree 4) their end behavior are top left bottom right 5) iin graph 3 bottom right going down and cross the y intercept -15

z(x) match graph 1 because
1) z(x) is in a standard form form i choose the first degree i take the highest exponent 2) positive leading coefficient odd degree. 3) end behavior bottom left top right. 4) there were two same equation on the intercept form

