# What are the vertex, focus and directrix of  y=x^2+10x+21 ?

Mar 4, 2016

Vertex is -5,-4), (focus is $\left(- 5 , - \frac{15}{4}\right)$ and directrix is $4 y + 21 = 0$

#### Explanation:

Vertex form of equation is $y = a {\left(x - h\right)}^{2} + k$ where $\left(h , k\right)$ is vertex

The given equation is $y = {x}^{2} + 10 x + 21$. It may be noted that the coefficient of $y$ is $1$ and that of $x$ too is $1$. Hence, for converting the same, we have to make terms containing $x$ a complete square i.e.

$y = {x}^{2} + 10 x + 25 - 25 + 21$ or

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

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

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

Standard form of parabola is ${\left(x - h\right)}^{2} = 4 p \left(y - k\right)$,

where focus is $\left(h , k + p\right)$ and directrix $y = k - p$

As the given equation can be written as ${\left(x - \left(- 5\right)\right)}^{2} = 4 \times \frac{1}{4} \left(y - \left(- 4\right)\right)$, we have vertex $\left(h , k\right)$ as $\left(- 5 , - 4\right)$ and

focus is $\left(- 5 , - \frac{15}{4}\right)$ and directrix is $y = - 5 - \frac{1}{4} = - \frac{21}{4}$ or $4 y + 21 = 0$