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

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Answer 1 · 2017-07-18T12:44:51

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

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

Comparing with standard form of vertex form equation

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

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

Vertex is at equidistance from focus and directrix and at opposite

sides . The distance of vertex from directrix is $d = 1/(4|a|) :.$

$d= 1/(4*4)=1/16$ . since $a>0$ , the parabola opens upwards and

directrix is below vertex. So directrix is $y= (-1-1/16)= -17/16=-1 1/16$

and focus is at $(3, (-1+1/16) )or (3,-15/16)$

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

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