# How do you find the maximum and minimum of y=(x+2)^2-3?

Maximum: $\infty$ , Minimum: $- 3$
$y = {\left(x + 2\right)}^{2} - 3 = {x}^{2} + 4 x + 1$ Comparing with quadratic equation $y = a {x}^{2} + b x + c$
Here $a = 1 , b = 4 , c = 1$.Discriminant $D = {b}^{2} - 4 a c = {4}^{2} - 4 \cdot 1 \cdot 1 = 12$
If $a > 0$ ; maximum is $\infty$ and minimum is $y = - \frac{D}{4 a} = - \frac{12}{4} = - 3$ graph{x^2+4x+1 [-10, 10, -5, 5]}[Ans]