How do you find the vertex and intercepts for #y=2(x+2)^2+3#?

1 Answer
Binayaka C.
Jun 4, 2017

Vertex is at #(-2,3)# , y-intercept is at #0,11# , No x-intercept.

Explanation:

#y=2(x+2)^2 +3 #. Comparing with standard vertex form of equation
#y=a(x-h)^2 +k ; (h,k)#being vertex , we find here #h=-2,k=3#
Hence vertex is at #(-2,3)#

y-intercept is obtained by putting #x=0# in the equation.
#:. y=2(0+2)^2+3 =11# or at # (0,11) #

x-intercept is obtained by putting #y=0# in the equation.
#:. 2(x+2)^2 +3=0 or 2(x+2)^2 = -3 or (x+2)^2 = -3/2# or
#(x+2) = +- sqrt(-3/2) or x = -2 +- sqrt(-3/2)# or
# x = -2 +- sqrt(3/2)i :. x# has complex roots.
So there is no x-intercept.

graph{2(x+2)^2+3 [-40, 40, -20, 20]} [Ans]

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