Question

In the following exercises, graph.

$\frac{{\left(y+3\right)}^2}{16}-\frac{{\left(x-3\right)}^2}{36}=1$

Short answer

The graph is as below

Step-by-step solution

Step 1. Given information

The equation is$\frac{{\left(y+3\right)}^2}{16}-\frac{{\left(x-3\right)}^2}{36}=1$

Draw the graph.

Step 2. Calculation

The y² term is positive, the transverse axis is vertical.

The hyperbola opens up and down.

Find the centre

Compare the equation with standard form $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$

So, $\left(h,k\right)=\left(3,-3\right)$and $a^2=16\Rightarrow a=\pm 4$, $b^2=36\Rightarrow b=\pm 6$

Sketch the rectangle that goes through the points 4 units above/below the centre 6 units left/right of centre.

Step 3. Draw the graph

Sketch the asymptotes and draw the graph.