How do you graph #y=(x-3)^2 +5#?

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Answer 1 · 2017-12-07T20:55:24

The graph should look like this: graph{(x-3)^2+5 [-10, 10, -5, 5]}

First, since the equation is in the vertex form #a(x-h)^2+k#, we find the a, h, and k.

In the function #y=(x-3)^2+5#, #a=1#, #h=3#, and #k=5#

Now, we plot the points.

We plot the points by using the following diagram:
Remember that when you plug in h, you always get k.
x======y
#h-2a#
#h-a#
#h# => #k#
#h+a#
#h+2a#

Plug these values in the function to get the y coordinates.

x======y
#3-2(1)#
#3-1#
#3#
#3+1#
#3+2(1)#

x==y
#1#=>#9#
#2#=>#6#
#3#=>#5#
#4#=>#6#
#5#=>#9#

Plot these points and you have the answer!