Question

Question #c5468

Answer 1

Extreme (minimum) value of the function `y` is at `x=(-1)/2`

Explanation:

We have `y=x^2+x+2`

As we can see `a>0` so its an upward facing parabola.
So it will have a minimum value.

The minumum value will be at `x=(-b)/(2a)`

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

                               OR

When we differentiate the function `y` we get

`y^'=2x+1`

When we find out the critical point of `y^'` it is

`2x+1=0`

`2x=-1`

`x=(-1)/2`

Therefore the function will be minimum at `x=(-1)/2`

When we put `x=(-1)/2` in `y` we get

`y=1/4-1/2+2`

`y=7/4`

`y=1.75`

Also the graph for the function is

graph{x^2+x+2 [-6.024, 6.46, -0.516, 5.724]}

Here you can see the minimum value of the function `y` is at `x=-0.5` which is `y=1.75`