How do you find the vertex and the intercepts for #y - 10 = 15 + x^2 + 2x#?

1 Answer
Feb 7, 2017

minimum at #(-1,24) # and y-intercept is #25#

Explanation:

#y-10 = 15+x^2+2x#

rearrange,
#y = x^2+2x+25#

y - intercept, when x=0,
#y = 25#

there is no x-intercept since #b^2-4ac < 0# where #a=1, b=2 and c=25#

by completing a square,
#y=(x+1)^2-(1)^2+25#
#y=(x+1)^2+24#
This is a minimum vertex because there is a +ve coefficient for #(x+1)^2#

therefore, it vertex is minimum at #(-1,24) # and y-intercept is #25#