Question

What is the vertex of the graph of `y = 2(x – 3)^2 + 4`?

Answer 1

Vertex is `(3,4)`

Explanation:

If the equation of parabola is of the form `y=a(x-h)^2+k`,

the vertex is `(h,k)`.

Observe that when `x=h`, the value of `y` is `k` and as `x` moves on either side, we have `(x-h)^2>0` and `y` rises.

Hence, we have a minima at `(h,k)`. It would be maxima if `a<0`

Here we have `y=2(x-3)^2+4`, hence we have vertex at `(3,4)`, where we have a minima.

graph{2(x-3)^2+4 [-6.58, 13.42, 0, 10]}

Answer 2

`"vertex "=(3,4)`

Explanation:

`"the equation of a parabola in "color(blue)"vertex form"` is.

`•color(white)(x)y=a(x-h)^2+k`

`"where "(h,k)" are the coordinates of the vertex and a"`
`"is a multiplier"`

`y=2(x-3)^2+4" is in this form"`

`"with "(h,k)=(3,4)larrcolor(magenta)"vertex"`

`"and "a=2`

`"since "a>0" then graph is a minimum"`
graph{2(x-3)^2+4 [-20, 20, -10, 10]}