Answer
See graph
Work Step by Step
We are given the hyperbola:
$\dfrac{(x+4)^2}{9}-\dfrac{(y+3)^2}{16}=1$
The equation is in the standard form:
$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$
The transverse axis is parallel to the $x$-axis.
Determine $h,k,a,b,c$:
$h=-4$
$k=-3$
$a^2=9\Rightarrow a=\sqrt 9=3$
$b^2=16\Rightarrow b=\sqrt {16}=4$
$c^2=a^2+b^2$
$c^2=9+16$
$c^2=25$
$c=\sqrt{25}$
$c=5$
Determine the centre of the hyperbola:
$(h,k)=(-4,-3)$
Determine the coordinates of the vertices:
$(h-a,k)=(-4-3,-3)=(-7,-3)$
$(h+a,k)=(-4+3,1)=(-1,-3)$
Determine the coordinates of the foci:
$(h-c,k)=(-4-5,-3)=(-9,-3)$
$(h+c,k)=(-4+5,-3)=(1,-3)$
Determine the asymptotes:
$y-k=\pm\dfrac{b}{a}(x-h)$
$y+3=\pm\pi$
Graph the hyperbola:
