Question

How do you graph the quadratic function and identify the vertex and axis of symmetry for `y=-(x-2)^2-1`?

Answer 1

`(2,-1),x=2`

Explanation:

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

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))`
where (h ,k) are the coordinates of the vertex and a is a constant.

`y=-(x-2)^2-1" is in this form"`

and by comparison `h=2" and " k=-1`

`rArr"vertex "=(2,-1)`

Since the `color(blue)"value of a is negative"`

`"That is " color(red)(-)(x-2)^2`

Then the parabola opens down vertically `color(red)(nnn`

The axis of symmetry goes through the vertex and therefore has equation `color(magenta)"x=2"`

`color(blue)"Intercepts"`

`x=0toy=-(0-2)^2-1=-4-1=-5`

`rArry=-5larrcolor(red)"y-intercept"`

`y=0to-(x-2)^2-1=0to(x-2)^2=-1`

This has no real solutions hence there are no x-intercepts.

Knowing the ' shape' of the parabola, the vertex and the y-intercept enables the graph to be sketched.
graph{-(x-2)^2-1 [-10, 10, -5, 5]}