How do you find the vertex and the intercepts for #f(x) = 4x^2 - 16x + 17#?

1 Answer
Apr 15, 2016

vertex: #(x,y)=(2,1)#
y-intercept: #17#
(no x-intercepts)

Explanation:

Rewriting the given equation: #f(x)=4x^2-16x+17#
as
#color(white)("XXX")f(x)=4(x^2-4x)+17#

#color(white)("XXX")f(x)=color(blue)(4)(x^2-4xcolor(red)(+4))+17color(blue)(-4)(color(red)(4))#

#color(white)("XXX")f(x)=4(x-2)^2+1#
which is the vertex form with vertex at #(2,1)#

The y-intercept is the value of #y# (i.e.#f(x)#) when #x=0#
#color(white)("XXX")f(0)=4(0-2)^2+1 =17#

The x-intercepts (if they exist) are the values of #x# for which #f(x)=0#
...but #(x-2)^2>=0# for all values of #x#
and therefore #4(x-2)^2+1 > 0# for all values of #x#.
#color(white)("XXX")f(x)!=0# for any value of #x#
graph{4x^2-16x+17 [-2, 9.096, -0.598, 4.95]}