Let's make a table of values to graph the given polynomial function. When you are making a table of values make sure to use a variety of points, including negative and positive values.
x
x^3+1
y=x^3+1
- 2
( -2)^3+1
-7
- 1
( -1)^3+1
0
0
0^3+1
1
1
1^3+1
1
2
2^3+1
9
We will now plot the obtained points and connect them with a smooth curve. Consider also that this is an odd-degreepolynomial with a positive leading coefficient.
&f(x) → -∞ as x → - ∞
&f(x) → + ∞ as x → + ∞
This tells us about the end behavior of the function.
We will now find the zeros of the given polynomial function. Recall that they are the points where a function intercepts the x-axis. To find all of the zeros, we can set the value of the function equal to zero and solve the obtained equation.
y = x^3 + 1
⇓
0 = x^3 + 1
We will solve the equation by factoring. To do so we will apply the sum of cubes.
a^3+b^3 ⇔ (a+b)(a^2-ab+b^2)
Let's factor the right-hand side of the equation using the formula.
From Equation (I) we found that one solution is x=-1. To find the other solutions we will solve Equation (II). Note that this is a quadratic equation. Therefore, we will use the Quadratic Formula.
ax^2+bx+c=0
⇕
x=- b±sqrt(b^2-4ac)/2a
To do so, we first need to identify a, b, and c.
x^2-x+1=0
⇕
1x^2+( - 1)x+ 1=0
We see that a= 1, b= - 1, and c= 1. Let's substitute these values into the formula and solve for x.
These solutions to the quadratic equation are also solutions for the polynomial equation. The solutions to the polynomial equations are also zeros of the given polynomial function. Note that two of them are complex numbers.
Zeros
x=- 1, x=1± isqrt(3)/2
b
Similarly as in Part A, let's make a table of values to graph the given polynomial function. When you are making a table of values make sure to use a variety of points, including negative and positive values.
x
x^3-8
y=x^3-8
- 2
( -2)^3-8
-16
- 1
( -1)^3-8
-9
0
0^3-8
-8
1
1^3-8
-7
2
2^3-8
0
We will now plot the obtained points and connect them with a smooth curve. Consider also that this is an odd-degree polynomial with a positive leading coefficient.
&f(x) → -∞ as x → - ∞
&f(x) → + ∞ as x → + ∞
This tells us about the end behavior of the function.
We will now find the zeros of the given polynomial function. Recall that they are the points where a function intercepts the x-axis. To find all of the zeros, we can set the value of the function equal to zero and solve the obtained equation.
y = x^3 - 8
⇓
0 = x^3 - 8
We will solve the equation by factoring. To do so, we will apply the difference of cubes.
a^3-b^3 ⇔ (a-b)(a^2+ab+b^2)
Let's factor the right-hand side of the equation using the formula.
From Equation (I), we found that one solution is x=2. To find other solutions we will solve Equation (II). Note that this is a quadratic equation. Therefore, we will use the Quadratic Formula.
ax^2+bx+c=0
⇕
x=- b±sqrt(b^2-4ac)/2a
To do so, we first need to identify a, b, and c.
x^2-2x+4=0
⇕
1x^2+( - 2)x+ 4=0
We see that a= 1, b= - 2, and c= 4. Let's substitute these values into the formula and solve for x.
These solutions to the quadratic equation are also solutions for the polynomial equation. The solutions to the polynomial equations are also zeros of the given polynomial function. Note that two of them are complex numbers.
Zeros
x=2, x=1± isqrt(3)
