How do you find the vertex of #y= x^2-8x+5#?

2 Answers
Aug 1, 2015

The simplest method for finding the vertex of the given polynomial is to convert it into "vertex form" to find the vertex at #(4,-11)#

Explanation:

Given #y=x^2-8x+5#

If we can convert this into vertex form:
#color(white)("XXXX")##y=m(x-a)^2+b#
#color(white)("XXXX")##color(white)("XXXX")#for some constants #a# and #b#
then the vertex will be at #(a,b)#

The easiest way to do this conversion is via a "completion of the square method"

If #x^2-8x# are the first 2 terms of a squared binomial
#color(white)("XXXX")##color(white)("XXXX")#note: #(m+n)^2 = (m^2+2mn+n^2)#
then the third term needs to be #((-8x)/(2x))^2 =16#

So we add (and then subtract #16# to "complete the square"

#y = (x^2-8x+16) +5 -16#

Simplifying:
#y =(x-4)^2 + (-11)#
#color(white)("XXXX")#which is our desired "vertex form" (with #m=1#)

So the vertex is at #(4,-11)#

Aug 1, 2015

Find vertex of f(x) = x^2 - 8x + 5

Ans: Vertex (4, -11)

Explanation:

x-coordinate of vertex: #x = -b/(2a) = 8/2 = 4#

y-coordinate of vertex: #y = f(4) = 4^2 - 8(4) + 5 = - 11#