Question

How do you graph `f(x)=(x+1)^2-2` and identify the x intercepts, vertex?

Answer 1

See below.

Explanation:

In the form `color(blue)(a)(x-color(red)(h))^2+color(green)(k)`. `color(blue)(a)` is the coefficient of `x^2`, `color(red)(h)` is the axis of symmetry and `color(green)(k)` is the maximum/minimum value of the function.

You already have this form, so:

Axis of symmetry `=color(red)(h)=-1`

Since the coefficient of `x^2` is positive the function will have a minimum value:

So:

Minimum value `=color(green)(k)= -2`

The vertex is `( color(red)(h) , color(green)(k)) = (-1 , -2)`

y axis intercept where `x=0`

`y= (0+1)^2-2=> y=-1`

`x` axis intercepts where `y=0`

`y=(x +1)^2-2=0=> (x+1)^2=2`

`x=-1+- sqrt(2)`

`(-1+sqrt(2) , 0 ) , (-1 - sqrt(2) , 0)`

Graph of `y=(x+1)^2-2`

graph{y= (x+1)^2-2 [-6.24, 6.243, -3.12, 3.12]}