Question

How do you graph `y = (x-2)^2 +2`?

Answer 1

`P_("vertex")->(x,y)=(2,2) =>`axis of symmetry is `x=2`

`P_("y-Intercept") ->(x,y)=(0,6)`

No `x"-intercept"`

The general shape of the graph is `uu`

Explanation:

If your question wording 'how do you graph' were to be taken literally you would get the following answer:

Draw up a table of values by substituting for `x` and calculate the related values of y. Mark off the points and draw the best fit that you can where the line passes through each marked point.
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Assumption: You intend the question to be: Determine the important points of `y=(x-2)^2+2` and sketch the graph.

Let any point on the graph be `P`

`color(blue)("Determine the general shape of the graph")`

Expanding the brackets of `y=+1(x-2)^2+2` we have:

`y=+x^2-4x+6`

As the coefficient of `x^2` is positive we have the general shape of `uu`. Thus the vertex is a minimum.

Note that if the coefficient had been negative then you would have had the general shape of `nn`

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`color(blue)("Determine the vertex")`

The given equation format is that of 'vertex form'

From this we can directly read off the vertex coordinates:

`color(blue)(P_("vertex")->(x,y)=(2,2))`
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`color(blue)("Determine the y-intercept")`

Substitute `x=0`

`y_("intercept")=(0-2)^2+2 = 6`

`color(blue)("P_("y-Intercept") ->(x,y)=(0,6)"`
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`color(blue)("Determine the x-intercept")`

Set `y=0`

`0=(x-2)^2+2`

Subtract 2 from both sides

`(x-2)^2=-2`

Square root both sides

`x-2=+-sqrt(-2)`

The square root of a negative number means that the graph does not cross the x-axis.

`color(brown)("If you required to have some actual values for x-intercept")`

`x=+2+-sqrt(2xx(-1))`

`x=+2+-sqrt(2)sqrt(-1)`

`x=+2+-sqrt(2)color(white)(.)i`

Where this is of complex number type of format:

'Real number' + 'Imaginary number' `-> Re+ Im`