Certainly! Let's work through the function step by step to define [tex]\( h(x) \)[/tex].
### Problem Statement
We are given the function [tex]\( h(x) \)[/tex] defined as:
[tex]\[ h(x) = 4x^3 + 2x^2 - 5 \][/tex]
### Breaking Down the Function
1. Cubic Term: The term [tex]\( 4x^3 \)[/tex] is a cubic term, meaning it has a degree of 3. This term will dominate the behavior of the function for very large or very small values of [tex]\( x \)[/tex].
2. Quadratic Term: The term [tex]\( 2x^2 \)[/tex] is a quadratic term, meaning it has a degree of 2. This term affects the curvature of the function but less so than the cubic term as [tex]\( x \)[/tex] grows larger in magnitude.
3. Constant Term: The term [tex]\(-5 \)[/tex] is a constant. This term shifts the entire function vertically downward by 5 units.
### Putting It All Together
The function [tex]\( h(x) \)[/tex] combines these three terms to create a unique polynomial of degree 3:
[tex]\[ h(x) = 4x^3 + 2x^2 - 5 \][/tex]
### Summary:
Therefore, for any value of [tex]\( x \)[/tex], [tex]\( h(x) \)[/tex] is calculated by plugging [tex]\( x \)[/tex] into the function [tex]\( 4x^3 + 2x^2 - 5 \)[/tex].
To illustrate, if we were to evaluate [tex]\( h(x) \)[/tex] at a specific point, for example, [tex]\( x = 1 \)[/tex]:
[tex]\[ h(1) = 4(1)^3 + 2(1)^2 - 5 \][/tex]
[tex]\[ h(1) = 4 + 2 - 5 \][/tex]
[tex]\[ h(1) = 1 \][/tex]
In conclusion, the function [tex]\( h(x) = 4x^3 + 2x^2 - 5 \)[/tex] describes a cubic polynomial with specific contributions from cubic, quadratic, and constant terms.
