How do you find the zero's for y=x^4-3x^2+2?

Question

And write a line tangent to the graph at x=1

Answer 1

u substitution
let u = x^2
y = x^4-3x^2+2 becomes
y = u^2 - 3u +2
to find zeros, you set the function equal to zero
0 = u^2 - 3u +2
0 = (u-2)(u-1)
0 = u - 2 and 0 = u -1
remember u = x^2, so
0 = x^ - 2 and 0 = x^ - 1
x^ = 2 and x^2 = 1
x = plus/minus sqrt 2 and x = plus/minus sqrt 1

Tangent at 1, so f(1) = 0, so we need tangent line through (1,0):
well first we need the slope of the function at 1, because that would be the slope of the tangent line.
so find derivative of the function and then evaluate it for x = 1
f'(x) = 4x^3-6x
f'(1) = 4-6 = -2
equation of line: y -y1 = m(x-x1)
y - 0 = -2(x-1)
y = -2x +2

Answer 2

One way to find the zeroes is to use the substitution z = x² and rewrite the polynomial in z as
y = z² - 3z +2, which may be factored as
y = (z - 1)(z - 2)
This you can back-substitute to get
y = (x² - 1)(x² - 2) which may then be further factored via differences of squares as
y = (x - 1)(x + 1)(x - √2)(x + √2)
giving zeroes at x = ±1 and x = ±√2.

The tangent line at (1, 0) - we know the y coordinate is zero because x = 1 is one of the zeroes that we just computed - we can differentiate y with respect to x, the slope of the line being the derivative of y at x = 1. This gives

y' = 4x^3 - 6x
= 4(1)^3 -6(1)
= 4 - 6
= -2

We now have a slope for the tangent line and a point through which it must pass, so we can think "point-slope form" to get
y - yo = m*(x - xo)
or
y - 0 = -2(x - 1)

Converting this to more convenient slope-intercept form gives the equation y = -2x + 2.

Answer 3

zeros: x^4 -3x^2 +2 = 0
Factor: (x^2 -1)(x^2 -2) = 0
set each factor equal to 0
x^2 - 1 = 0 or x^2 - 2 = 0
x^2 = 1 or x^2 = 2 take the square root
x = +/- 1 or x = +/- sq rt of 2
zeros: 1, -1. sqrt2, -sqrt2

To find the tangent line's slope, you find the derivative of y and then evaluate when x = 1
y' = 4x^3 -6x
when x = 1 this is -2
Since 1 was a zero, then a point on the graph is (1,0)

With the slope of -2 and the point (1,0)
the equation of the tangent line is found by using the point-slope formula
y - 0 = -2(x - 1)
y = -2x + 2

Answer 4

You don't have examples in your book? Always look at those and figure out what they did.

You need to factor your equation, then set each factor equal to zero and solve for x.

Do you remember "F.O.I.L."? you mutiply "First, Outside, Inside, Last" To factor, you kind of do a reverse FOIL. What two numbers when combined/added together will give you -3 and also multiply to give you a positive 2? ... -1 and -2

What two numbers will mutiply together to give you x4 .... x2 and x2

so you've got: x4 - 3x2 + 2 = ( x2 - 1 ) ( x2 - 2 )
(use FOIL to see if that is correct . . . and it is ... so now you can find your zeros)

x2 - 1 = 0 ... goes to: x2 = 1 ... therefore x can be +1 or -1
and
x2 - 2 = 0 ... goes to x2 = 2 ... therefore x can be + or - sqare root of 2

so you've got four possible answers. Are these correct? Try them out ...
first one is +1
go back and plug it in for each x in your original equation

(1)4 - 3(1)2 + 2 . . . = . . . 1 - 3(1) + 2 . . . 1-3+2 .... -2+2 = 0
yep!
try the rest, if they all work the same, then they are all zeros of the originial equation

Answer 5

y = x^4 - 3x^2 + 2

y = x^4 - 2x^2 - x^2 + 2

=>x^2(x^2-2) - 1(x^2-2)

=>(x^2-1)(x^2-2) =>(x+1)(x-1)(x^2-2)

so zeros of the given polynomial are 1, -1, +/-sqrt(2)

b)
the slope of tangent is derivative of y

y' = 4x^3 - 6x

at x = 1

y'(1) = 4 - 6 = -2

the slope of tangent = -2

the point of tangency = (1, y(1))

to find out y(1) substitute x = 1 in y

y(1) = 1 - 3 + 2 = 0

so point of tangency = (1,0)

the equation of tangent wirh slope -2 and going through (1,0)

is

y-0 = -2(x-1)

y = -2x + 2

Answer 6

Put x^2 = z so
z^2-3z+2=0
z= ((3+-sqrt(9-8))/2
so z= 2 z=1
so x= +-sqrt(2) and x= +-1
y´=4x^3-6x ´and at x= 1 is =-2
so the tangent is
y= -2(x-1)
y= -2x+2

Answer 7

How about trying to factor

y = x^4 - 3x^2 +2
y = (x^2 -2)(x^2-1) ; this gives you the 0s for y
x^2 = 2, 1
x = +-sqrt2 and +-1

Hope that helps.

Answer 8

Substitute t = x^2. Then
t^2 -3t +2 = (t-2)(t-1)
so t =2 and t = 1
x^2 = 2 and x^2 =1
x = +/- sqrt 2 and x = +/- 1

Answer 9

have you tried the rational zero test