Graphing F(x) = (x-3)(x+1)²: A Step-by-Step Guide

Hey guys! Let's dive into graphing the function f(x) = (x-3)(x+1)². It might look a bit intimidating at first, but trust me, we'll break it down step-by-step. Understanding how to graph polynomial functions like this is super useful in math, science, and even real-world applications. We’ll explore key features like intercepts, multiplicity of roots, and the overall behavior of the graph. So, buckle up and let's get started!

Understanding the Function f(x) = (x-3)(x+1)²

Before we even think about drawing lines and curves, let's get to know our function a little better. The function we're dealing with is f(x) = (x-3)(x+1)². This is a polynomial function, which means it's made up of terms with variables raised to non-negative integer powers. In our case, we have a factored form, which is super helpful because it directly tells us about the roots of the function. Roots, also known as x-intercepts, are the points where the graph crosses the x-axis. They are the values of x for which f(x) equals zero. To find the roots, we simply set each factor equal to zero and solve for x. So, we have (x-3) = 0 and (x+1)² = 0. Solving these, we find that x = 3 and x = -1 are our roots. But here's where it gets interesting: notice the square on the (x+1) factor? This tells us something about the multiplicity of the root. The root x = -1 has a multiplicity of 2 because the factor (x+1) is squared. The root x = 3 has a multiplicity of 1 because the factor (x-3) is raised to the power of 1 (which we don't usually write, but it's there!). Multiplicity affects how the graph behaves at the x-intercept. A root with a multiplicity of 1, like x = 3, means the graph will simply pass through the x-axis at that point. However, a root with a multiplicity of 2, like x = -1, means the graph will touch the x-axis and then turn around, kind of like a bounce. This is a crucial detail that will help us sketch the graph accurately. We also need to consider the degree of the polynomial. The degree is the highest power of x in the polynomial. To find the degree, we could expand the function, but there's a quicker way. We can simply add up the powers of x in each factor. In our case, we have x from the (x-3) factor (which is x to the power of 1) and x² from the (x+1)² factor. So, the degree of the polynomial is 1 + 2 = 3. The degree tells us about the end behavior of the graph, which is what happens to the function as x approaches positive or negative infinity. For a polynomial with an odd degree (like 3), the graph will have opposite end behaviors: it will either go up on the left and down on the right, or down on the left and up on the right. To determine which one, we look at the leading coefficient, which is the coefficient of the term with the highest power of x. In our case, if we were to expand the function, the leading term would be x³ (1 * x * x²). The coefficient of x³ is 1, which is positive. This means the graph will go down on the left (as x approaches negative infinity) and up on the right (as x approaches positive infinity). Knowing all these details – the roots, their multiplicities, the degree, and the leading coefficient – gives us a solid foundation for sketching the graph.

Finding Key Features: Intercepts and Behavior

Alright, let's get down to the nitty-gritty of finding the key features that will help us sketch this graph. We've already touched on the roots, but let's solidify that and then move on to other important points. As we discussed, the roots, or x-intercepts, are the points where the graph crosses or touches the x-axis. We found these by setting f(x) = 0 and solving for x. This gave us x = 3 and x = -1. So, we have two x-intercepts: (3, 0) and (-1, 0). Remember, x = -1 has a multiplicity of 2, meaning the graph will bounce off the x-axis at this point, while x = 3 has a multiplicity of 1, meaning the graph will pass straight through the x-axis. Another crucial feature to find is the y-intercept. The y-intercept is the point where the graph crosses the y-axis. This occurs when x = 0. To find it, we simply plug in x = 0 into our function: f(0) = (0-3)(0+1)² = (-3)(1)² = -3. So, the y-intercept is (0, -3). This gives us another key point to plot on our graph. Now, let's think about the behavior of the graph between these intercepts. We know the graph goes down on the left and up on the right, and we know it crosses the x-axis at x = 3 and bounces off the x-axis at x = -1. We also know it crosses the y-axis at y = -3. This gives us a general idea of the shape of the graph, but to get a more accurate picture, we can consider some additional points. We could plug in a few more x-values into our function and see what y-values we get. For example, we could try x = 1: f(1) = (1-3)(1+1)² = (-2)(2)² = -8. So, we have the point (1, -8). This tells us that the graph dips down quite a bit between x = -1 and x = 3. Similarly, we could try x = -2: f(-2) = (-2-3)(-2+1)² = (-5)(-1)² = -5. So, we have the point (-2, -5). This tells us that the graph dips down a bit to the left of x = -1 before it starts to rise again. These additional points help us refine our sketch and get a better sense of the curve of the graph. To get an even more precise graph, we could use calculus to find the local maximum and local minimum points, which are the turning points of the graph. These are the points where the graph changes direction, going from increasing to decreasing or vice versa. However, for a basic sketch, the intercepts and the general behavior are usually enough. By finding the x and y-intercepts and considering the behavior of the graph based on the roots and the leading coefficient, we've gathered enough information to create a reasonable sketch. Remember, the more points you plot, the more accurate your graph will be!

Sketching the Graph of f(x) = (x-3)(x+1)²

Okay, we've done the groundwork, now it's time for the fun part: actually sketching the graph! Grab your paper and pencil (or your favorite graphing software) and let's bring this function to life. We're going to use all the information we've gathered so far to create an accurate representation of f(x) = (x-3)(x+1)². First things first, let's plot the key points we've identified. We have the x-intercepts at (3, 0) and (-1, 0), and the y-intercept at (0, -3). Go ahead and mark these points on your graph. Remember that at x = -1, the graph will bounce off the x-axis because of the multiplicity of 2, while at x = 3, it will pass straight through. This is a crucial detail to keep in mind as we draw the curve. We also found the additional points (1, -8) and (-2, -5), which will help us refine the shape of the graph. Plot these points as well. Now, let's think about the end behavior. We know that the graph goes down on the left (as x approaches negative infinity) and up on the right (as x approaches positive infinity) because the leading coefficient is positive and the degree is odd. This gives us the general direction the graph will take as it moves away from the center. Starting from the left side of the graph, we know it's going to come from down below. As it moves towards the x-axis, it will eventually reach the point (-1, 0). Since this is a root with a multiplicity of 2, the graph will touch the x-axis and then turn around, heading back down. This creates a sort of