Question

Helpppppppppppppppp please

Answer 1

• y=x⁴+3x²-x-3

As there is the highest degree 4 there are four zeros .

Using calculator

the roots are(attached)

The graph is also attached

Answer 2

Answer:

a) 4 roots

b) x = 1 and x = -3

$\textsf{c)} \quad x=1, \quad x=-3, \quad x=\dfrac{-1 + \sqrt{3}i}{2}, \quad x=\dfrac{-1 - \sqrt{3}i}{2}$

d) see attached

Step-by-step explanation:

Given function:

$f(x)=x^4+3x^3-x-3$

Part (a)

From inspection of the function, the highest exponent of the polynomial is 4.  Therefore, it is expected that the polynomial will have 4 roots.

Part (b)

To find the rational roots, find values of x where $f(x)=0$

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

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

Therefore, the rational roots are: x = 1 and x = -3

Part (c)

As there are only 2 rational roots, the other 2 roots must be complex roots. To find these, first factor the polynomial using the 2 rational roots found in part (b):

$\implies f(x)=(x-1)(x+3)(x^2+bx+1)$

$\implies f(x)=(x^2+2x-3)(x^2+bx+1)$

$\implies f(x)=x^4+bx^3+x^2+2x^3+2bx^2+2x-3x^2-3bx-3$

$\implies f(x)=x^4+(2+b)x^3+(2b-2)x^2+(2-3b)-3$

Compare coefficients to find b:

$\implies (2+b)x^3=3x^3$

$\implies b=1$

Therefore the fully factored function is:

$\implies f\:\!(x)=(x-1)(x+3)(x^2+x+1)$

Therefore:

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

To find the complex roots, use the quadratic formula

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

$\implies x=\dfrac{-1 \pm \sqrt{1^2-4(1)(1)}}{2(1)}$

$\implies x=\dfrac{-1 \pm \sqrt{-3}}{2}$

$\implies x=\dfrac{-1 \pm \sqrt{3}\sqrt{-1}}{2}$

$\implies x=\dfrac{-1 \pm \sqrt{3}i}{2}$

Therefore, all the roots of the function are:

$x=1, \quad x=-3, \quad x=\dfrac{-1 + \sqrt{3}i}{2}, \quad x=\dfrac{-1 - \sqrt{3}i}{2}$

Part (d)

End behaviors:

As the degree of the function is even and the leading coefficient is positive, the end behaviors are:

$\textsf{As } x \rightarrow - \infty, \:\: f(x) \rightarrow + \infty$

$\textsf{As } x \rightarrow + \infty, \:\: f(x) \rightarrow + \infty$

y-intercept

Substitute x = 0 into the function:

$f\:\!(0)=(0)^4+3(0)^3-(0)-3=-3$

Therefore, the y-intercept is (0, -3)

To find the approximate x-coordinates of the turning points (stationary points), differentiate the function, set it to zero, then solve for x:

$f'\:(x)=4x^3+9x^2-1$

$\implies 4x^3+9x^2-1=0$

Using a calculator, x ≈ -2.2, x ≈ -0.4, x ≈ 0.3

Therefore the approximate coordinates of the turning points are:

(-2.2, -9.3),  (-0.4, -2.7)  and (0.3, -3.2)

We don't need to plot these points, they merely help us with the approximate shape of the curve.