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

1 Answer
Jan 19, 2017

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.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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))
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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)"
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

Tony BTony B