How do you find the vertex and intercepts for #y = (1/8)(x – 5)^2 - 3#?

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Answer 1 · 2018-04-07T11:53:19

Vertex is at #(5,-3) # y intercept is #y=1/8 # and
x intercepts are at ** #(5+ 2sqrt6 ,0) and(5- 2sqrt6 ,0)#

#y=1/8(x-5)^2-3# Comparing with vertex form of

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

here #h=5 , k=-3 :.# Vertex is at #(5,-3) #, y intercept

is found by putting #x=0# in the equation #y = 1/8(0-5)^2-3# or

#y=25/8-3 or y= 1/8 :.# y intercept is #y=1/8# or at

# (0,1/8)#, x intercept is found by putting #y=0#

in the equation #:.0 = 1/8(x-5)^2-3 # or

#1/8(x-5)^2=3 or (x-5)^2 = 24 or (x-5)= +-sqrt24 # or

#x= 5+- 2sqrt6 #, x intercepts are at** #(5+ 2sqrt6 ,0) #

and#(5- 2sqrt6 ,0)#

graph{1/8(x-5)^2-3 [-45, 45, -22.5, 22.5]} [Ans]