Asked by Dan
Without graphing the function f(x)=(2x+1)(x-4)^2 determine:
a) the degree of the function
b) the type of polynomial function
c) the leading coefficient
d) the zeroes
e) the y-intercept
f) the end behaviors
a) the degree of the function
b) the type of polynomial function
c) the leading coefficient
d) the zeroes
e) the y-intercept
f) the end behaviors
Answers
Answered by
oobleck
(a) x^1 * x^4 = x^5
(b) type? we have the degree, so what else is there?
I guess you could say "with real coefficients"
(c) 2*1^4 = ___
(d) if (x-a) is a factor, the x=a is a zero
(e) odd degree, with positive leading coefficient, so with x outside the interval containing all the roots, y<0 when x<0 and y>0 when x>0
(b) type? we have the degree, so what else is there?
I guess you could say "with real coefficients"
(c) 2*1^4 = ___
(d) if (x-a) is a factor, the x=a is a zero
(e) odd degree, with positive leading coefficient, so with x outside the interval containing all the roots, y<0 when x<0 and y>0 when x>0
Answered by
mathhelper
expand it to see that the highest power is x^3, so the degree is 3
so it is a cubic function
the leading coefficient comes from the (2x+1) , so it is 2
the zeros are -1/2 and 4
for the y-intercept, let x = 0, then evaluate
for large values of x, (x-4)^2 is always +large, so is the 2x+1, so +large
for large negative values of x, (x-4)^2 is still +larg, but 2x+1 becomes negative lare, so - large
so it is a cubic function
the leading coefficient comes from the (2x+1) , so it is 2
the zeros are -1/2 and 4
for the y-intercept, let x = 0, then evaluate
for large values of x, (x-4)^2 is always +large, so is the 2x+1, so +large
for large negative values of x, (x-4)^2 is still +larg, but 2x+1 becomes negative lare, so - large
Answered by
oobleck
thanks, mathhelper
you were somewhat more careful than I was ...
you were somewhat more careful than I was ...
Answered by
Dan
Thank you both!
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