Question

Find the extreme values of `y= x^2 − 2x + 3`?

Answer 1

Minimum at `x=1` maximum at `+oo` and `-oo`

Explanation:

To get the minimum set the derivative `d/dx(x^2-2x+3)=2x-2`
equal to zero:
`2x-2=0`
which solves for `x=1`
Since the highest exponent in this equation is 1 it only has one solution.
By doing the second derivative `d/dx(2x-2)=2` you can see at x=1 its positive therby its a minimum.
The maximums are, since the highest exponent in the original equation (2) is even and has a positive factor, at `+oo` and `-oo`

Answer 2

Solution part 1 of 2: All answer for `x`; part answer for `y`

Upper bound `->x=+-oo and y=+oo`

Explanation:

The term `x^2` 'grows' faster than the other term involving `x`

So to use a none mathematical term; `x^2` wins.

So `lim_(x->oo) y=[lim_(x->oo)x^2]-[2lim_(x->oo)x]+3`

Tends to

`lim_(x->oo) y=[lim_(x->oo)x^2]`

Note that if `x<0` then `x^2>0`
Also that if `x>0` then `x^2>0` as well

Thus we write:

`lim_(x->oo) y=[lim_(x->oo)x^2] -> k` where `k=+oo`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Foot note

As the `x^2` term is positive then the general shape is that of type `uu`. So the `+oo` matches that context.

Answer 3

Solution part 2 of 2: The whole and final answer.

`-oo < x < +oo`

`color(white)("dd.")2 <=y < +oo`

Explanation:

Given: `y=x^2-2x+3`

`x` can and may take on the value of `0`

As the `x^2` term is positive then the general shape of the curve is that of `uu`. Thus there is a minimum for `y`

To determine this we need the vertex (bottom of the `uu`)

This will be `1/2` way between the x-intercepts if there are any.
Looking at the given equation notice that `(-1)xx(-3)=+3`

But `-1-3 !=-2` thus it is more straight forward to use the formula or you can do this sort of cheat. It is part of the various steps to complete the square.

Given
`y=ax^2+bx+c` write as `y=a(x+b/(2a))^2+k`

`x_("vertex") =(-1)xxb/(2a) = (-1) xx (-2)/(2xx1)=+1`

Substitute `x=1` into the original equation.

`y_("vertex")=(1)^2-2(1)+3 = +2`

Vertex `->(x,y)=(1,2)`