How do you graph #y = x^2 - 6x +11#?

1 Answer
David O. · Stefan V.
Apr 22, 2018

Find the vertex, then find another point.

Explanation:

To graph the equation #x^2-6x+11# by hand, it is easiest to find a) the vertex, and b) one additional point. The vertex can be found by using these steps:

#x_"coordinate" = -b/(2a)#

for

#y=ax^2+bx+c#

So

#x_"coordinate" = -(-6)/(2 * 1) = 3#

Vertex: #(3, y)#

Then

#(3)^2-6(3)+11=2#

So

Vertex: #(3, 2)#

For the next point, use any #x# integer value, i.e. #5#

#(5)^2-6(5)+11=6#

Point: #(5, 6)#

Now, use these points(and some more if you want to) to graph your equation. Remember, all points on one side of the vertex can be reflected to the other side. [(5, 6) and (-5, 6)]

Also, graphing calculators such as desmos.com/calculator may also be of help to you. Try it out!

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