Graph f(x)=x^2-2x

Problem

ƒ(x)=x2−2*x

Solution

1. Identify the function type and its properties. The function ƒ(x)=x2−2*x is a parabola that opens upward because the leading coefficient is positive (a=1.

2. Find the vertex using the formula x=(−b)/(2*a) For this function, a=1 and b=−2

x=(−(−2))/(2*(1))=1

ƒ(1)=(1)2−2*(1)=−1

The vertex is at (1,−1)

3. Find the x-intercepts by setting ƒ(x)=0 and factoring.

x2−2*x=0

x*(x−2)=0

The x-intercepts are at (0,0) and (2,0)

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

ƒ(0)=0−2*(0)=0

The y-intercept is at (0,0)

5. Plot additional points to refine the shape. For x=−1

ƒ*(−1)=(−1)2−2*(−1)=3

For x=3

ƒ(3)=(3)2−2*(3)=3

6. Sketch the graph by drawing a smooth U-shaped curve through the points (−1,3) (0,0) (1,−1) (2,0) and (3,3)

Final Answer

ƒ(x)=x2−2*x

(The graph is a parabola with vertex at (1,−1) and x-intercepts at x=0 and x=2

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