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

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Answer 1 · 2015-12-18T06:44:22

The vertex is #(4,2)#.

#y=x^2-8x+18# is a quadratic equation in the form #ax^2+bx+c#, where #a=1, b=-8, and c=18#.

The vertex is the maximum or minimum point of a parabola.

To find the #x# value of the vertex use the formula #x=(-b)/(2a)#.

#x=(-(-8))/(2*1)#

#x=8/2#

#x=4#

To find the value of #y#, substitute the value for #x# into the equation.

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

#y=4^2-(8*4)+18#

#y=16-32+18#

#y=2#

The vertex is #(4,2)#.

graph{y=x^2-8x+18 [-9.944, 10.06, -0.07, 9.93]}