How do you find the vertex and the intercepts for #f(x) = (x + 5)^2 - 2#?

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Answer 1 · 2018-05-29T04:35:58

Vertex is at #(-5, -2) # x intercepts are at
# (-5+sqrt 2, 0) and (-5-sqrt 2, 0)#, y intercept is at #(0,23)#

# f(x) =(x+5)^2-2 # . Comparing with vertex form of

equation #f(x) = a(x-h)^2+k ; (h,k)# being vertex we find

here #h=-5 , k= -2 :.# Vertex is at #(-5, -2) # . y intercept

is found by putting #x=0# in the equation #y = (x+5)^2-2# or

#y=(0+5)^2-2 or y =23 :.# y intercept is #y=23 or (0,23)#

x intercept is found by putting #y=0# in the equation

#y = (x+5)^2-2 or 0 = (x+5)^2-2 or (x+5)^2=2#

#(x+5)=+- sqrt 2 or x = -5+- sqrt 2 :.# x intercepts are

at # (-5+sqrt 2, 0) and (-5-sqrt 2, 0)#.

graph{(x+5)^2-2 [-45, 45, -22.5, 22.5]}