How do you graph #18x ^ { 4} - 3x ^ { 2} - 1= 0#?

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Answer 1 · 2018-02-08T13:16:11

#x approx +- 0.577#

#18x^4-3x^2-1=0#

Since the question is an equation we will first solve for #x#

Let #phi =x^2#

#:. 18phi^2 -3phi-1=0#

#(3phi-1)(6phi+1)=0#

#phi = 1/3 or -1/6#

Since, #x =+-sqrt(phi): x in RR -> phi >=0#

#:. x = +-sqrt(1/3) approx +- 0.577#

Now, to produce a graph.

To graph the function #y=18x^4-3x^2-1# we can note the following:

#y# has zeros at #approx +- 0.577#

#y=-1# at #x=0#

Either by plotting points or by using calculus we can find that #y# has a minimum value of #-1.125# at #x approx +-0.289#.

#y# has no finite upper bounds.

The graph of #y# is shown below.

graph{18x^4-3x^2-1 [-1.513, 1.525, -1.302, 0.217]}