Certainly! Let's derive the cubic polynomial step-by-step given the conditions provided.
1. Identify the x-intercepts and their behavior:
- The function has x-intercepts at [tex]\(x = 2\)[/tex] and [tex]\(x = 4\)[/tex].
- The function "bounces" at [tex]\(x = 4\)[/tex], which means the polynomial touches the x-axis at this point but does not cross it. Therefore, [tex]\(x = 4\)[/tex] is a double root.
2. Construct the polynomial form:
- Given the x-intercepts and bouncing behavior, we can write the polynomial as:
[tex]\[
p(x) = a(x - 2)(x - 4)^2
\][/tex]
Here, [tex]\(a\)[/tex] is a constant that needs to be determined.
3. Use the point (6, 8) to determine [tex]\(a\)[/tex]:
- We are given that the polynomial goes through the point [tex]\((6, 8)\)[/tex]. This provides us with the equation p(6) = 8.
- Substitute [tex]\(x = 6\)[/tex] and [tex]\(p(x) = 8\)[/tex] into the polynomial equation:
[tex]\[
8 = a(6 - 2)(6 - 4)^2
\][/tex]
Simplify inside the parentheses:
[tex]\[
8 = a(4)(2)^2
\][/tex]
[tex]\[
8 = a(4)(4)
\][/tex]
[tex]\[
8 = 16a
\][/tex]
- Solve for [tex]\(a\)[/tex]:
[tex]\[
a = \frac{8}{16}
\][/tex]
[tex]\[
a = \frac{1}{2}
\][/tex]
4. Write the final polynomial:
- Substitute [tex]\(a = \frac{1}{2}\)[/tex] back into the polynomial form:
[tex]\[
p(x) = \frac{1}{2}(x - 2)(x - 4)^2
\][/tex]
Therefore, the cubic polynomial with x-intercepts at 2 and 4, where the function bounces at 4 and passes through the point (6, 8), is:
[tex]\[ p(x) = \frac{1}{2}(x - 2)(x - 4)^2 \][/tex]
This is the required polynomial.
