Question

How to graph a parabola `y=1/2(x-3)^2+5`?

Answer 1

Please read the explanation.

Explanation:

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`color(green)("Step 1"`

Construct a data table with input `color(red)(x` and corresponding values for `color(red)y`:

This table will help immensely in understanding the End Behavior of the given function: `color(blue)(y=f(x)=(1/2)*(x-3)^2+5`

`color(red)(x: -5<=x<=5` [ Col 1 ]

Draw graphs for `y=x^2`, `y=(x-3)^2`, `y=(1/2)(x-3)^2`, and finally `y=(1/2)(x-3)^2+5`

Find Vertices, x-intercept and y-intercept, if any, for all the graphs.

`color(green)("Step 2"`

`color(red)("Graph: " y=x^2` .....Parent Quadratic Function

Useful to analyze the End-behavior of quadratic functions.

`color(green)("Step 3"`

`color(red)("Graph: " y=(x-3)^2`

`color(green)("Step 4"`

`color(red)("Graph: " y=(1/2)*(x-3)^2`

`color(green)("Step 5"`

`color(red)("Graph: " y=(1/2)*(x-3)^2+5`

`color(green)("Step 6"`

View all the graphs together:

What we observe ?

`color(brown)(y=f(x)=(1/2)*(x-3)^2+5`

1. General form: `color(blue)(y=f(x)=a(x-h)^2+k`, Vertex: `color(green)((h.k)`

2. On the graph in `color(green)("Step 4")` we have `color(blue)(Vertex: (3,5)`

3. Graph Opens up, as the `x^2` term is positive.

4. Parabolic curve is expanded outward, as `color(red)(0 < a < 1`

5. `x=h`, and in our problem `x=3` is the Axis of Symmetry

6. `h=3` indicates the Horizontal Shift

7. `k=5` indicates the Vertical Shift