Question

Make a complete graph of the following functions. A graphing utility is useful in locating intercepts, local extreme values, and inflection points. $$f(x)=3 x^{4}+4 x^{3}-12 x^{2}$$

Short answer

Answer: The key features of the given function include intercepts, local extreme values (maximum and minimum), and inflection points. The x-intercept and y-intercept are both at (0,0), there is a local maximum at x=0 and a local minimum at x=1, and there are no inflection points. These key features help create an accurate graph by providing points and information about the function's behavior, making it easier to sketch the shape of the polynomial function.

Step-by-step solution

Find the intercepts

To find the x-intercepts, set the function equal to zero and solve for x: $$3x^4 + 4x^3 - 12x^2 = 0$$ Factor out the common factor of $$x^2$$: $$x^2(3x^2 + 4x - 12) = 0$$ Now, solve for x: $$x = 0$$ is an x-intercept. For the other factor, $$3x^2 + 4x - 12$$, you may use a graphing calculator or apply the quadratic formula. You'll find that there are no other real roots. Therefore, the x-intercept is $$x = 0$$. To find the y-intercept, set x = 0: $$f(0) = 3(0)^4 + 4(0)^3 - 12(0)^2 = 0$$ The y-intercept is $$y = 0$$.

Find the local extreme values

To find the local extreme values, we need to find the critical points of the function. To do this, find the first derivative and set it equal to zero: $$f'(x) = 12x^3 + 12x^2 - 24x$$ Factor out the common factor of $$12x$$: $$12x(x^2 + x - 2) = 0$$ Now, solve for x: $$x = 0$$ Again, use a graphing calculator or apply the quadratic formula to find other critical points in the equation $$x^2 + x - 2$$: $$x = \frac{-1 \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)}$$ $$x = \frac{-1 \pm \sqrt{9}}{2}$$ $$x = -2, 1$$ To determine whether these critical points are local maxima or minima, use the second derivative test. Find the second derivative of the function: $$f''(x) = 36x^2 + 24x - 24$$ Now, evaluate the second derivative at each critical point: $$f''(-2) = 72 - 48 - 24 = 0$$ $$f''(0) = -24 < 0$$ $$f''(1) = 36 + 24 - 24 = 36 > 0$$ According to the second derivative test, the function has a local maximum at $$x=0$$ and a local minimum at $$x=1$$.

Find the inflection points

To find the inflection points, we need to determine where the second derivative changes sign. Set the second derivative equal to zero and solve for x: $$36x^2 + 24x - 24 = 0$$ This is a quadratic equation, so you may use a graphing calculator or apply the quadratic formula to find the other critical points: $$x = \frac{-24 \pm \sqrt{(-24)^2 - 4(36)(-24)}}{2(36)}$$ $$x = \frac{-24 \pm \sqrt{2304}}{72}$$ However, there is no real solution for x. Thus, there are no inflection points.

Create the graph

Now that we have all the key features, use a graphing utility to make the graph. Plot the x-intercept and y-intercept (0,0), the local maximum at $$x=0$$, the local minimum at $$x=1$$, and note that there are no inflection points. With this information, create a smooth curve that goes through the plotted points and shows the local maximum and minimum, capturing the shape of the polynomial function.

Key concepts

Finding Intercepts

Intercepts of a polynomial function are essential in understanding its graph. X-intercepts, also known as zeros or roots, occur where the function crosses or touches the x-axis (where the output of the function, f(x), equals 0). To find these, we set the function equal to zero and solve for x. For our polynomial function, $$f(x)=3x^4+4x^3-12x^2$$, factoring out an $$x^2$$ leaves us with $$x^2(3x^2 + 4x - 12) = 0$$. After breaking it down, we discover the x-intercept at $$x = 0$$.
The Y-intercept is where the graph crosses the y-axis (where the input, x, is 0). Setting x to 0 in our function, we find that the y-intercept is also at the point (0, 0). Finding these intercepts helps sketch the foundation of the graph.

Local Extreme Values

Local extreme values — the hills and valleys on the graph of a function — signify where a function reaches a minimum or maximum within a certain interval. To locate these on our polynomial function, we employ the first derivative test. We first find the derivative, $$f'(x) = 12x^3 + 12x^2 - 24x$$, and set it to zero to find potential local extremes. Factoring out $$12x$$, we then solve for x: $$x = 0, -2, 1$$. By further analyzing these critical points with the second derivative, we determined that our function has a local maximum at $$x = 0$$ and a local minimum at $$x = 1$$. This step is pivotal as it enables us to map out the peaks and dips of the graph.

Inflection Points

Inflection points are where the graph of a function changes concavity from 'smiling' (concave up) to 'frowning' (concave down) or vice versa. These are found where the second derivative of the function changes sign. For our function, we look to $$f''(x) = 36x^2 + 24x - 24$$ and set it to zero to find potential points of inflection. Solving this quadratic equation might usually give us the x-values of these points, but in this case, there are no real solutions. Therefore, our graph has no inflection points, which suggests that the concavity does not change.

First Derivative Test

The first derivative test is used to determine whether a critical point is a local maximum or minimum. After finding that $$x = 0, -2, 1$$ from the first derivative of our function, we then evaluate the second derivative at each critical point. We find $$f''(-2) = 0$$, $$f''(0) = -24$$, and $$f''(1) = 36$$. The negative second derivative at $$x = 0$$ indicates a local maximum, and the positive value at $$x = 1$$ indicates a local minimum. By applying the first derivative test, we can understand not just the location but also the nature of critical points on the graph of our polynomial function, helping us to accurately sketch the high and low points.