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

2 Answers
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)}

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