Graph y=(x-3)^2+2

Problem

y=(x−3)2+2

Solution

1. Identify the form of the equation. This is a quadratic function in vertex form, y=a*(x−h)2+k where (h,k) is the vertex of the parabola.

2. Determine the vertex by identifying h and k from the expression (x−3)2+2

h=3

k=2

Vertex=(3,2)

3. Determine the direction of the opening. Since the coefficient a=1 is positive, the parabola opens upward.

4. Find the y-intercept by substituting x=0 into the equation.

y=(0−3)2+2

y=9+2

y=11

y-intercept=(0,11)

5. Find additional points to define the shape. Choose xvalues near the vertex, such as x=2 and x=4

For *x=2:y=(2−3)2+2=1+2=3

For *x=4:y=(4−3)2+2=1+2=3

Points=(2,3),(4,3)

6. Sketch the graph by plotting the vertex (3,2) the y-intercept (0,11) and the points (2,3) and (4,3) then drawing a smooth U-shaped curve through them.

Final Answer

y=(x−3)2+2

The graph is a parabola opening upward with vertex at (3,2) and y-intercept at (0,11)

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