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

1 Answer
Dec 4, 2015

This is already in vertex form making it easy to identify the vertex ...

Explanation:

vertex form: #y=a(x-h)^2+k#, with vertex #(h,k)#

vertex #=(3,1)#

The vertex is above the x-axis and since the coefficient #a=2# is positive, this parabola opens upward and will have no x-intercepts (only 2 imaginary roots).

The y-intercept happens when #x=0#

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

hope that helped

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