How do you find the vertex and intercepts for #y = x^2 - 8x + 18#?

2 Answers
Özgür Özer
Nov 23, 2015

Vertex={(4,2)}, intercept={(0,18)}

Explanation:

#y=x^2-8x+18#
#=>y'=2x-8#
For #y'=0#,
#2x-8=0#
#=>x=4#
#=>y=4^2-8*4+18#
#=16-32+18#
#=2#
#=> vertex={(4,2)}#
#Delta=(-8)^2-4*1*18#
#=-8#
#=> Delta<0#
For #y=0#,
#x^2-8x+18=0#
#=># #x# is imaginary
For #x=0#,
#y=18#
#=># intercept={(0,18)}

Alan P.
Nov 23, 2015

vertex: #(4,2)#
y-intercept: #18#
there is no x-intercept

Explanation:

Converting the given equation into vertex form: #y=m(x-a)^2+b# with vertex at #(a,b)#

#y=x^2-8x+18#

#rarr y=x^2-8x+16+2#

#rarr y=1(x-4)^2+2#
#color(white)("XXX")with vertex at #(4,2)#

y-intercept is the value of #y# when #x=0#
#rArr# y-intercept = 18#

x-intercept is the value of #x# when #y=0#
i.e. when #x^2-8x+18=0#
but checking the discriminant (#b^2-4ac# using the standard form)
we see that there are no solutions for #x# since the discriminant is #< 0#

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