Sketch the graph: (x+2)^2/25-(y+4)^2/25=1

I don't understand this at all! When I tried graphing it I just had two lines going through the center, which is (-2,-4). But it should be a hyperbola, right? Could anyone help me?

hyperbola

center at (-2,-4)
eccentricity = √2
asymptotes y+4 = ±(x+2)
vertices at x = ±2.5
foci at x = ±5√2

Oops

vertices at (-2±5,-4)

So do I draw lines through the vertices?

Of course! I can help you understand how to sketch the graph of the equation (x+2)^2/25 - (y+4)^2/25 = 1, which represents a hyperbola.

To sketch the graph of a hyperbola, it's helpful to know its standard form:

((x - h)^2/a^2) - ((y - k)^2/b^2) = 1

In this case, the center of the hyperbola is at (-2, -4) because of the expression (x + 2)^2 and (y + 4)^2. The values inside the parentheses indicate the horizontal and vertical shifts of the hyperbola's center, respectively.

Additionally, a^2 and b^2 represent the horizontal and vertical distances from the center to the vertices of the hyperbola. In this case, both a^2 and b^2 are equal to 25. Thus, a and b are both equal to 5.

Knowing this information, we can start sketching the graph by plotting the center at (-2, -4). Then, we can locate the vertices by moving horizontally a distance of a (5) units from the center.

From the center, move 5 units to the right to mark the vertex V1 at (3, -4). Then, move 5 units to the left to mark the vertex V2 at (-7, -4).

Next, locate the conjugate axis endpoints by moving vertically a distance of b (5) units from the center.

From the center, move 5 units upward to mark the conjugate axis endpoint C1 at (-2, 1). Then, move 5 units downward to mark the conjugate axis endpoint C2 at (-2, -9).

Now you can sketch the asymptotes of the hyperbola passing through the center. The asymptotes are lines that intersect at the center of the hyperbola. In this case, the asymptotes have slopes of ±(b/a) = ±1, meaning they pass through the points (h, k) = (-2, -4). This gives us the equations y = x - 2 (positive slope) and y = -x - 6 (negative slope).

Finally, you can draw the curves of the hyperbola, making sure they remain symmetric with respect to the center.

With all this information, you should be able to sketch the graph of the given hyperbola.