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

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Answer 1 · 2018-04-01T04:54:28

vertex is #(3,1)#

Y intercept 19
and
No x intercept

In vertex form #f(x)=A(B[x-C])^2+D#

We know that C is the x co-ordinate of the vertex and D is the y co-ordinate

So the vertex is #(3,1)#

Y intercept (when x 0)

#y=2((0)-3)^2+1=2(-3)^2+1=18+1=19#

X intercept (when y 0)

#0=2(x-3)^2+1#
#-1=2(x-3)^2#
#sqrt(-1)=2(x-3)#

Root 1 doesn't exist on the number line showing that there is no x intercept