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

Jun 23, 2018

Minimum at $x = 1$ maximum at $+ \infty$ and $- \infty$

#### Explanation:

To get the minimum set the derivative $\frac{d}{\mathrm{dx}} \left({x}^{2} - 2 x + 3\right) = 2 x - 2$
equal to zero:
$2 x - 2 = 0$
which solves for $x = 1$
Since the highest exponent in this equation is 1 it only has one solution.
By doing the second derivative $\frac{d}{\mathrm{dx}} \left(2 x - 2\right) = 2$ 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 $+ \infty$ and $- \infty$

Jun 23, 2018

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

Upper bound $\to x = \pm \infty \mathmr{and} y = + \infty$

#### Explanation:

The term ${x}^{2}$ 'grows' faster than the other term involving $x$

So to use a none mathematical term; ${x}^{2}$ wins.

So ${\lim}_{x \to \infty} y = \left[{\lim}_{x \to \infty} {x}^{2}\right] - \left[2 {\lim}_{x \to \infty} x\right] + 3$

Tends to

${\lim}_{x \to \infty} y = \left[{\lim}_{x \to \infty} {x}^{2}\right]$

Note that if $x < 0$ then ${x}^{2} > 0$
Also that if $x > 0$ then ${x}^{2} > 0$ as well

Thus we write:

${\lim}_{x \to \infty} y = \left[{\lim}_{x \to \infty} {x}^{2}\right] \to k$ where $k = + \infty$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Foot note

As the ${x}^{2}$ term is positive then the general shape is that of type $\cup$. So the $+ \infty$ matches that context.

Jun 23, 2018

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

$- \infty < x < + \infty$

$\textcolor{w h i t e}{\text{dd.}} 2 \le y < + \infty$

#### Explanation:

Given: $y = {x}^{2} - 2 x + 3$

$x$ 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 $\cup$. Thus there is a minimum for $y$

To determine this we need the vertex (bottom of the $\cup$)

This will be $\frac{1}{2}$ way between the x-intercepts if there are any.
Looking at the given equation notice that $\left(- 1\right) \times \left(- 3\right) = + 3$

But $- 1 - 3 \ne - 2$ 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
$y = a {x}^{2} + b x + c$ write as $y = a {\left(x + \frac{b}{2 a}\right)}^{2} + k$

${x}_{\text{vertex}} = \left(- 1\right) \times \frac{b}{2 a} = \left(- 1\right) \times \frac{- 2}{2 \times 1} = + 1$

Substitute $x = 1$ into the original equation.

${y}_{\text{vertex}} = {\left(1\right)}^{2} - 2 \left(1\right) + 3 = + 2$

Vertex $\to \left(x , y\right) = \left(1 , 2\right)$