Problem 35 - College Algebra

Question

Use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$\frac{(x+3)^{2}}{25}-\frac{y^{2}}{16}=1$$

Step-by-Step Solution

Answer

The center of the hyperbola is (-3,0), the vertices are at (-8,0) and (2,0), and the foci are at (-3 ± √41,0). The equations of the asymptotes are \( y = \pm \frac{4}{5}(x+3) \).

Step 1: Identify the center of the hyperbola

The center of the hyperbola is given by the values inside the parentheses in the equation and with opposite signs. So, the center is at (-3,0).

Step 2: Identify the vertices of the hyperbola

The vertices are located at a distance of \( \sqrt{25} = 5 \) units to the left and right of the center along the x-axis. So, the vertices are at (-8,0) and (2,0).

Step 3: Identify and sketch the asymptotes

The asymptotes of the hyperbola are given by the lines passing through the center with slopes of \( \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5} \). The equations of the lines are \( y = \pm \frac{4}{5}(x+3) \). Plot these lines on the graph.

Step 4: Sketch the hyperbola

Draw the hyperbola, making sure it approaches the asymptotes as \( x \) goes to \( \pm \infty \). The hyperbola passes through the vertices and opens to the left and right since the x term is positive in the equation.

Step 5: Find the foci of the hyperbola

The distance from the center to each focus is given by \( \sqrt{25+16} = \sqrt{41} \) units. So the foci are at \( (-3 \pm \sqrt{41},0) \).