# What are the vertex, focus and directrix of  y=4(x-3)^2-1 ?

Jul 18, 2017

Vertex is at $\left(3 , - 1\right)$ , focus is at $\left(3 , - \frac{15}{16}\right)$ and
directrix is $y = - 1 \frac{1}{16}$.

#### Explanation:

$y = 4 {\left(x - 3\right)}^{2} - 1$

Comparing with standard form of vertex form equation

y= a (x-h)^2+k ; (h,k) being vertex , we find here

$h = 3 , k = - 1 , a = 4$. So vertex is at $\left(3 , - 1\right)$.

Vertex is at equidistance from focus and directrix and at opposite

sides . The distance of vertex from directrix is $d = \frac{1}{4 | a |} \therefore$

$d = \frac{1}{4 \cdot 4} = \frac{1}{16}$. since $a > 0$, the parabola opens upwards and

directrix is below vertex. So directrix is $y = \left(- 1 - \frac{1}{16}\right) = - \frac{17}{16} = - 1 \frac{1}{16}$

and focus is at $\left(3 , \left(- 1 + \frac{1}{16}\right)\right) \mathmr{and} \left(3 , - \frac{15}{16}\right)$

graph{4(x-3)^2-1 [-10, 10, -5, 5]} [Ans]