Question

PLEASE HELP!
Sketch the graph of the function by a) applying the Leading Coefficient Test, b) finding the zeros of the polynomial, c) plotting sufficient solution points, and d) drawing a continuous curve through the points.

g(x)=x^4-4x^2

Answer 1

a) applying the Leading Coefficient Test,
g(x)= x^4-4x^2
^
+1
b) finding the zeros of the polynomial,
g(x)=x^4-4x^2 set y=0
0=x^4-4x^2 solve for "x"

Answer 2

for a what is the +1 for? give me just a second to work out b!

Answer 3

the "leading coefficient test" just means.... check the number in front of the leading term
if the number is positive, the graph goes one way
if the number is negative, it turns upside down

Answer 4

for b i got x=0 x=2 x=-2

Answer 5

is that correct?

Answer 6

yes \(\bf y=x^4-4x^2\implies 0=x^2(x-2)(x+2)\)

notice the "x squared", it means \(\bf x^2=0\implies x\cdot x=\implies x=0, \quad x=0\)

is has a "multiplicity" of 2, when the root multiplicity is even, it means at that point, the graph HITS the x-axis BUT DOESN'T CROSS IT, just touches it and backs up/down

Answer 7

now you'd just need to ... plot a couple of points

like the middle x-value between -2 and 0 and 0 and 2
to see where the "hump" will be, either up or down

Answer 8

how do i know which ppints to plot?

Answer 9

well... between x-intercepts, for a quadratic... the middle point between them gives you the vertex

for a quartic, like this one... it might be a bit off, but close enough

so the middle or close to middle or maybe even 3, the middle value between the x-intercepts and a couple more to the side of it will do

Answer 10

fwhat about (-1,-4) (-2,4)(0,0)(1,-4)(2,4)? could those be points?

Answer 11

well... -2 is a root.... so, that'd be -2,0

more like something like .... -1.5, -1, -0.5
+1.5, +1, +0.5

Answer 12

\(\bf y=x^4-4x^2\qquad x=\{\pm1.5, \pm1, \pm0.5\}\)

Answer 13

so then those are what the points should be?? im sorry im a bit confused on how you got that