Graph f(x)=(x-5)^2

Problem

ƒ(x)=(x−5)2

Solution

1. Identify the parent function as y=x2 which is a standard parabola opening upwards with its vertex at (0,0)

2. Determine the horizontal shift by looking at the term (x−5) Since the constant 5 is subtracted from x inside the square, the graph of y=x2 is shifted 5 units to the right.

3. Locate the vertex of the parabola. The shift moves the vertex from (0,0) to (5,0)

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

ƒ(0)=(0−5)2

ƒ(0)=25

The y-intercept is at (0,25)

5. Plot additional points to define the shape. For example, if x=4 or x=6

ƒ(4)=(4−5)2=1

ƒ(6)=(6−5)2=1

This gives the points (4,1) and (6,1)

6. Sketch the smooth U-shaped curve passing through the vertex (5,0) and the points (4,1) (6,1) and (0,25)

Final Answer

ƒ(x)=(x−5)2* is a parabola with vertex at *(5,0)

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