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

a) 4 roots

b) x = 1 and x = -3

`\textsf{c)} \quad x=1, \quad x=-3, \quad x=(-1 + √(3)i)/(2), \quad x=(-1 - √(3)i)/(2)`

d) see attached

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=(-b \pm √(b^2-4ac) )/(2a)\quad\textsf{when }\:ax^2+bx+c=0`

`\implies x=(-1 \pm √(1^2-4(1)(1)))/(2(1))`

`\implies x=(-1 \pm √(-3))/(2)`

`\implies x=(-1 \pm √(3)√(-1))/(2)`

`\implies x=(-1 \pm √(3)i)/(2)`

Therefore, all the roots of the function are:

`x=1, \quad x=-3, \quad x=(-1 + √(3)i)/(2), \quad x=(-1 - √(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.