How do you find the vertex and intercepts for #x^2-4x+y+3=0#?

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Answer 1 · 2018-06-08T19:24:19

See graph

#x^2-4x+y+3=0#

first put in standard form:

#y=-x^2+4x-3#

y-intercept, set #x=0# and solve for #y#

#y=0^2+4(0)-3#

#y=-3#

x-intercept(s) if they exist, set #y=0# and solve for #x#

#0=-x^2+4x-3#

factor:

#0=-(x - 1)(x - 3)#

#x=1# and #x=3#

Axis of Symmetry (aos) in the form #ax^2+bx+c#

#aos = (-b)/(2a)#

#y=-x^2+4x-3#

#aos = (-4)/(2(-1))#

#aos=2#

Finally find the vertex it is a minimum if #a>0# a maximum if #a<0#

#vertex = (aos, f(aso))#

#vertex = (2, f(2))#

#vertex = (2, -2^2+4*2-3)#

#vertex = (2, 1)# is a maximum.

graph{-x^2+4x-3 [-7.21, 12.79, -8.2, 1.8]}