# How do you find the vertex and the intercepts for y = (x + 5)^2 +2?

Mar 19, 2017

vertex: $\left(- 5 , 2\right)$
no $x$-intercepts
$y$-intercept: $\left(0 , 27\right)$

#### Explanation:

Standard form used for graphing a parabola: $y = a {\left(x - h\right)}^{2} + k$,
where vertex:$\left(h , k\right)$

Finding the vertex:
Since the function $y = {\left(x + 5\right)}^{2} + 2$ is in standard form, you can easily find the vertex to be $\left(- 5 , 2\right)$.

Finding $x -$intercepts by letting $y = 0$:
1. Distribute the function: $y = {x}^{2} + 5 x + 5 x + 25 + 2$
2. Add like terms: $y = {x}^{2} + 10 x + 27$
3. Factor or use the quadratic formula $x = \frac{- B \pm \sqrt{{B}^{2} - 4 A C}}{2 A}$
$x = \frac{- 10 \pm \sqrt{100 - 4 \cdot 1 \cdot 27}}{2} = \frac{- 10 \pm \sqrt{- 8}}{2} = - 5 \pm \sqrt{2} i$

No real solutions, only complex solutions. This means the function does not touch the $x -$axis. No $x -$intercepts

Finding $y$-intercepts: by letting $x = 0$:
$y = {5}^{2} + 2 = 27$
$y$-intercept: $\left(0 , 27\right)$