# How do you find the axis of symmetry, graph and find the maximum or minimum value of the function y= (x+1)^2?

Apr 4, 2016

$\textcolor{b l u e}{\text{Axis of symmetry is } x = - 1}$

$\textcolor{b l u e}{\text{It is a minimum } \to \left(x , y\right) \to \left(- 1 , 0\right)}$

#### Explanation:

Consider standard equation form $y = a {x}^{2} + b x + c$

Your equation is in standard vertex form of $y = a {\left(x + \frac{b}{2 a}\right)}^{2} + c$

Where $a = 1 \text{ and } c = 0$ and ${x}_{\text{vertex}} = \left(- 1\right) \times \frac{b}{2 a}$

${x}_{\text{vertex}} = \left(- 1\right) \times \left(1\right) = - 1$. This is also the axis of symmetry.

$\textcolor{b l u e}{\text{So axis of symmetry is } x = - 1}$

$\textcolor{b r o w n}{\text{The coefficient of "x" is +1 ie positive. So the graph is of}}$$\textcolor{b r o w n}{\text{general shape "uu", thus we have a minimum.}}$

${x}_{\text{minimum")=x_("vertex}} = - 1$

Thus by substitution

y_("minimum")=(x_("vertex")+1)^2

${y}_{\text{minimum}} = {\left(- 1 + 1\right)}^{2} = 0$

$\textcolor{b l u e}{\text{So the minimum } \to \left(x , y\right) \to \left(- 1 , 0\right)}$ 