Question

Graph each polynomial function. $$ f(x)=4 x^{4}+7 x^{2}-2 $$

Short answer

The graph of the polynomial \( f(x) = 4x^4 + 7x^2 - 2 \)rises to infinity on both ends and has a minimum at the y-intercept (0, -2).

Step-by-step solution

- Identify the Degree and Leading Coefficient

Identify the degree and leading coefficient of the polynomial function. The given function is \[ f(x) = 4x^4 + 7x^2 - 2 \]The polynomial has a degree of 4, which means it is a quartic function, and the leading coefficient is 4.

- Determine the End Behavior

Determine the end behavior using the degree and leading coefficient. For a quartic function with a positive leading coefficient, the end behavior is that both ends of the graph will rise to infinity. As \( x \rightarrow -\infty \), \( f(x) \rightarrow \infty \) and as \( x \rightarrow \infty \), \( f(x) \rightarrow \infty \).

- Find the Y-intercept

Substitute 0 for x to find the y-intercept of the polynomial function. \( f(0) = 4(0)^4 + 7(0)^2 - 2 = -2 \). So, the y-intercept is (0, -2).

- Analyze the Critical Points

To determine critical points, find the first derivative of the function and set it to zero. Find \( f'(x) = 16x^3 + 14x \). Set \( 16x^3 + 14x = 0 \).Factor the derivative to get \( 2x(8x^2 + 7) = 0 \). This equation gives critical points at \( x = 0 \) and the solutions to \( 8x^2 + 7 = 0 \) , which are complex, so we focus on \( x = 0 \).

- Use the Second Derivative Test

Find the second derivative to determine the concavity at the critical point. \( f''(x) = 48x^2 + 14 \). Substitute \( x = 0 \) into \( f''(x) \) to find \( f''(0) = 14 \), which is positive, indicating a local minimum at \( x = 0 \).

- Analyze the Graph

Combine the intercept, critical points, and end behavior to sketch the graph. The graph passes through (0, -2) with a local minimum at the origin, and rises to infinity on both ends.

Key concepts

Degree of a Polynomial

Understanding the degree of a polynomial is crucial in graphing functions correctly. The degree of a polynomial is determined by the highest power of the variable in the polynomial. For the given function \[ f(x) = 4x^4 + 7x^2 - 2 \] the highest power of \[ x \] is 4.
This makes it a quartic polynomial. Quartic means the polynomial has a degree of four.
The degree informs us about how many roots the polynomial can have and its general shape.
As with any polynomial, the degree also tells us the maximum number of turning points the graph can have. For a degree-four polynomial, it can have up to three turning points.

Leading Coefficient

The leading coefficient of a polynomial function is the coefficient of the term with the highest degree. In the function \[ f(x) = 4x^4 + 7x^2 - 2 \] the leading coefficient is 4.
This coefficient affects the width and the direction of the graph's ends. If it is positive, the ends of the graph will rise towards infinity; if it is negative, the ends will fall towards negative infinity. The magnitude of this coefficient influences how 'steep' the graph looks.

End Behavior

End behavior describes how the values of \[ f(x) \] behave as \[ x \] approaches positive or negative infinity. For the function \[ f(x) = 4x^4 + 7x^2 - 2 \] which is a quartic polynomial with a positive leading coefficient, both ends of the graph will rise towards infinity. As \[ x \rightarrow -fty \] and as \[ x \rightarrow fty \] \[ f(x) \rightarrow fty \]. This means that as \[ x \] becomes very large positive or very large negative, \[ 4x^4 \] grows much faster than any lower-degree term, causing the graph to rise upwards.

Y-Intercept

The y-intercept is the point at which the graph of the polynomial crosses the y-axis. To find it, we set \[ x \] to 0 in the function \[ f(x) \].
For our polynomial \[ f(x) = 4x^4 + 7x^2 - 2 \] when \[ x = 0, f(0) = -2 \] Therefore, the y-intercept is (0, -2).
This point provides a specific location that helps in sketching the graph.

Critical Points

Critical points occur where the first derivative of the function is zero or undefined. These points are important as they indicate local minima, maxima, or saddle points. We first find \[ f'(x) = 16x^3 + 14x \] Then, set the derivative to zero: \[16x^3 + 14x = 0\right \] \[ 16x^3 + 14x = 0 \] factors to \[ 2x(8x^2 + 7) = 0 \].
This gives critical points at \[ x = 0 \] as the solutions to \[ 8x^2 + 7 = 0 \] are complex.

Second Derivative Test

The second derivative test helps to determine the concavity of the graph at critical points. This indicates whether a critical point is a local minimum, maximum, or saddle point. For the given function, the second derivative is \[ f''(x) = 48x^2 + 14 \] When \[ x = 0 \] substituting it into the second derivative gives \[ f''(0) = 14 \]. Since this value is positive, it indicates a local minimum at \[ x = 0 \]. The graph is concave up at this point, confirming a local minimum.