Question

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

Answer 1

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]