Graph f(x)=(x-1)(x+3)^2

Problem

ƒ(x)=(x−1)*(x+3)2

Solution

1. Identify the x-intercepts by setting ƒ(x)=0 The roots are x=1 and x=−3

2. Determine the multiplicity of each root. The root x=1 has multiplicity 1, so the graph crosses the x-axis. The root x=−3 has multiplicity 2, so the graph touches the x-axis and turns around (a local extremum).

3. Find the y-intercept by evaluating ƒ(0)

ƒ(0)=(0−1)*(0+3)2

ƒ(0)=(−1)*(9)=−9

4. Determine the end behavior by looking at the leading term. Expanding the expression gives a leading term of x3 As x→∞ ƒ(x)→∞ As x→−∞ ƒ(x)→−∞

5. Find the local extrema by taking the derivative ƒ(x)′ using the product rule.

ƒ(x)′=1⋅(x+3)2+(x−1)⋅2*(x+3)

ƒ(x)′=(x+3)*(x+3+2*x−2)

ƒ(x)′=(x+3)*(3*x+1)

6. Solve for critical points by setting ƒ(x)′=0

(x+3)*(3*x+1)=0

x=−3,x=−1/3

7. Calculate the coordinates of the local maximum and minimum. We already know ƒ*(−3)=0 For x=−1/3

ƒ*(−1/3)=(−1/3−1)*(−1/3+3)2

ƒ*(−1/3)=(−4/3)*(8/3)2=−256/27≈−9.48

8. Sketch the graph using the end behavior (starts bottom-left, ends top-right), the turning point at (−3,0) the y-intercept at (0,−9) the local minimum at (−1/3,−256/27) and the x-intercept at (1,0)

Final Answer

ƒ(x)=(x−1)*(x+3)2

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