How do you use the important points to sketch the graph of #y=x^2-4x+5#?
1 Answer
May 20, 2016
Please see below.
Explanation:
An equation of the type
- The first thing is to identify axis of symmetry, which in above form is given by
#x=-b/2a# . Here we have#y=x^2-4x+5# and hence axis of symmetry is#x=(-(-4))/2=2# . - Further, as coefficient of
#x^2# is positive and at#x=2# #y=2^2-4*2+5=1# , it has a minima at#(2,1)# - At
#x=0# ,#y=5# hence it passes through#(0,5)# - Other points may be found around
#x=2# and these are#(-3,26)# ,#(-2,17)# ,#(-1,10)# ,#(1,2)# ,#(3,2)# ,#(4,5)# ,#(5,10)# ,#(6,17)# ,#(7,26)# .
The parabola looks like
graph{x^2-4x+5 [-7.375, 12.625, -1.8, 8.2]}
