Sketch the graph of the ff
1. f(x) = (x+4)(x-5)(x-1)
2. f(x) = −(x+1)²(x−1)³(x-3)
(x+2)³
3. f(x) = x²(x-2)³(x−1)(x+1)²
4. f(x) = x³(x-3)2(x+2)³(x-1)
(x+1)²

please i need this by tomorrow, answer it nicely

Let's break down how to sketch the graphs of these functions. We'll use the concept of "zeros" (where the graph crosses the x-axis) and the behavior of the function near those zeros to guide our sketching. General Approach 1. Find the Zeros: Set the function equal to zero and solve for x. These are the x-intercepts of the graph.2. Determine Multiplicity: The power of each factor tells you the multiplicity of the zero.- Odd Multiplicity: If the power of a factor is odd (like (x-2)³), the graph crosses the x-axis at that zero.- Even Multiplicity: If the power of a factor is even (like (x-1)²), the graph touches the x-axis at that zero but doesn't cross.3. End Behavior: Look at the leading term of the function (the term with the highest power of x).- Odd Degree: If the leading term has an odd degree, the graph will rise on one side and fall on the other.- Even Degree: If the leading term has an even degree, the graph will rise on both sides or fall on both sides.4. Sketch: Connect the dots using the information about zeros, multiplicity, and end behavior. You can also plot a few additional points if needed to get a better sense of the curve. Let's apply this to your functions: 1. f(x) = (x+4)(x-5)(x-1) - Zeros: x = -4, x = 5, x = 1 (all multiplicity 1, so the graph crosses the x-axis at each)- End Behavior: Leading term is x³, so it's odd degree. The graph will rise on the right and fall on the left.- Sketch: [asy]unitsize(0.4 cm); real func (real x) {return((x + 4)(x - 5)(x - 1));} draw(graph(func,-6,6),red);draw((-6,0)--(6,0));draw((0,-10)--(0,10)); dot((-4,0), red);dot((5,0), red);dot((1,0), red);[/asy] 2. f(x) = −(x+1)²(x−1)³(x-3) / (x+2)³ - Zeros: x = -1 (multiplicity 2, touches but doesn't cross), x = 1 (multiplicity 3, crosses), x = 3 (multiplicity 1, crosses), x = -2 (vertical asymptote, the graph approaches infinity as x approaches -2)- End Behavior: Leading term is -x⁶ (even degree). The graph will fall on both sides.- Sketch: [asy]unitsize(0.4 cm); real func (real x) {return(-(x + 1)^2 * (x - 1)^3 * (x - 3) / (x + 2)^3);} draw(graph(func,-5,-2.2),red);draw(graph(func,-1.8,-0.8),red);draw(graph(func,-0.6,1.8),red);draw(graph(func,2,4),red);draw((-5,0)--(4,0));draw((0,-10)--(0,10));draw((-2,-10)--(-2,10),dashed); dot((-1,0), red);dot((1,0), red);dot((3,0), red);[/asy] 3. f(x) = x²(x-2)³(x−1)(x+1)² - Zeros: x = 0 (multiplicity 2, touches but doesn't cross), x = 2 (multiplicity 3, crosses), x = 1 (multiplicity 1, crosses), x = -1 (multiplicity 2, touches but doesn't cross)- End Behavior: Leading term is x⁷ (odd degree). The graph will rise on the right and fall on the left.- Sketch: [asy]unitsize(0.4 cm); real func (real x) {return(x^2 * (x - 2)^3 * (x - 1) * (x + 1)^2);} draw(graph(func,-2.5,2.5),red);draw((-2.5,0)--(2.5,0));draw((0,-10)--(0,10)); dot((0,0), red);dot((2,0), red);dot((1,0), red);dot((-1,0), red);[/asy] 4. f(x) = x³(x-3)²(x+2)³(x-1) / (x+1)² - Zeros: x = 0 (multiplicity 3, crosses), x = 3 (multiplicity 2, touches but doesn't cross), x = -2 (multiplicity 3, crosses), x = 1 (multiplicity 1, crosses), x = -1 (vertical asymptote, the graph approaches infinity as x approaches -1)- End Behavior: Leading term is x⁶ (even degree). The graph will rise on both sides.- Sketch: [asy]unitsize(0.4 cm); real func (real x) {return(x^3 * (x - 3)^2 * (x + 2)^3 * (x - 1) / (x + 1)^2);} draw(graph(func,-3,-1.2),red);draw(graph(func,-0.8,-0.2),red);draw(graph(func,0.2,1.8),red);draw(graph(func,2,4),red);draw((-3,0)--(4,0));draw((0,-10)--(0,10));draw((-1,-10)--(-1,10),dashed); dot((0,0), red);dot((3,0), red);dot((-2,0), red);dot((1,0), red);[/asy]

Answered by aiahbinilat | 2024-10-16