Question

Let f be the function defined by f(x) =x^4 -3x^2 +2
A)find the zeros of f
B)write an equation of the line tangent to the graph of f at the point where x=1
C) find the x coordinate of each point at which the line tangent to the graph of f is parallel to the line y=-2x+4

Answer 1

The zeros of the function can be determined by equating the equation to zero and determining the values of x. 
 0 = x4  - 3x2 + 2 the roots are x1 = sqrt 2 x2 = -sqrt 2x3 = 1x4 = -1
The tangent line is determined by differentiating the polynomial and substituting x by 1 to get the slope. That is,  m  = 4x^3 - 6 x = 4*1 - 6 = -2y-y1 = m(x-x1)when x =1 , y = 1-3+2 = 0 
y - 0 = -2*(x-1)y = -2x + 2
c. y = -2x + 4  m = -2 -2 = 4x3 - 6x  x= -1.3666 ; y =-0.1160 x= 1.3666 ; y =--0.1160
x =1 ; y = 0

Answer 2

A convenient way to find the zeros of  $f(x)=x^4-3x^2 +2$ is by factoring.

a) The equation,

$x^4-3x^2 +2=0$

can be rewritten as,

$(x^2)^2-3(x^2) +2=0$

We can think of this equation as a quadratic equation in  $x^2$, with $a=1,b=-3\:\: and \:\: c=2$.

Observe that $ac=1 \times 2=2$.

We find two factors of 2 that adds up to $-3$. These are, $-2,-1$.

Now let us split the middle term. to obtain;

$(x^2)^2-(x^2) -2(x^2)+2=0$

We can factor to get,

$(x^2)(x^2-1)-2((x^2-1)=0$

We factor further to obtain;

$(x^2-1)((x^2-2)=0$

$\Rightarrow (x-1)(x+1((x- \sqrt{2})(x+ \sqrt{2})=0$

Hence the zeroes of $f(x)$ are;

$x=1,x=-1,x= \sqrt{2},x=- \sqrt{2}$

b) To find the line tangent, we must first, find the slope using differentiation. That is,

$Slope\:\: function=f'(x)=4x^3-6x$

At $x=1$,

$Slope=f'(1)=4(1)^3-6(1)=-2$

Also, we need to determine the $y$ value at $x=1$. That is;

$f(1)=(1)^4-3(1)^2+2=0$

Now we can use the slope $m=-2$ and the point $(1,0)$ to write ythe equation of the line tangent.

$y-y_1 =m(x-x_1)$

$\Rightarrow y-0 =-2(x-1)$

$\Rightarrow y=-2x+2$

c)

If the line tangent is parallel to the line $y=-2x+4$, then

$f'(x)=-2$

Since parallel lines have the same slope.

$\Righarrow 4x^3-6x=-2$

$\Righarrow 4x^3-6x+2=0$

$\Righarrow (x-1)(2x- ( \sqrt{3}-1)(2x+ ( \sqrt{3}-1)=0$

Hence the x-coordinates are,

$x=1,x= \frac{ \sqrt{3}-1}{2},x= -\frac{\sqrt{3}-1}{2}$