Question

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

Answer 1

The vertex is #(2,3). #X-intercepts are `2+-sqrt3.`
Y-intercept = -1.

Explanation:

Let `y=f(x)=3-(x-2)^2`
`:. y-3= -(x-2)^2`

Clearly, the vertex is `(2,3)`

To find intercepts on X-axis & Y-axis, we have to take `y=0, x=0` resp.

`y=0` `rArr -3=-(x-2)^2` `rArr` `(x-2)=+-sqrt3`
`rArr` `x=2+-sqrt3`

So, X-intercepts are `2+-sqrt3`

`x=0` `rArr` `y-3=-(0-2)^2 = -4`, so `y=-1` is Y-intercept.