# How do you find the vertex, the y-intercept, and symmetric point, and use these to sketch the graph given y=x^2+8x+21?

Nov 20, 2016

See the answer below

#### Explanation:

Let's write the equation in the standard vertex form.

$y = {x}^{2} + 8 x + 21 = {x}^{2} + 8 x + 16 + 5$

$y = {\left(x + 4\right)}^{2} + 5$

The vertex is $\left(- 4 , 5\right)$

The symmetric line is $x = - 4$

When $x = 0$, $\implies$, $y = 21$

The intercept with the y axis is $\left(0 , 21\right)$

There are no intercept with the x-axis.

graph{x^2+8x+21 [-30.5, 27.22, -2, 26.88]}