Question

Find the intercepts, if any, of the graph of x^2+y^2+4x-2y-11=0

Answer 1

Hello there. To solve this question, we'll have to remember some properties about finding the intercepts of a graph.

Given the function

x² + y² + 4x - 2y - 11 = 0

In this case, we want to determine the x, y intercepts, it is, when y = 0, x = 0, respectively.

Making y = 0, we find the x-intercepts

x² + 4x - 11 = 0

Solving this quadratic equation,

`x=\frac{-4\pm\sqrt[]{4^2-4\cdot1\cdot(-11)}}{2\cdot1}=\frac{-4\pm\sqrt[]{60}}{2}=\frac{-4\pm2\sqrt[]{15}}{2}=-2\pm\sqrt[]{15}`

The intercepts are (-2 - sqrt(15), 0) and (-2 + sqrt(15), 0)

Now, to find the y-intercepts, plug in x = 0:

y² -2y - 11 = 0

Same process, solving this quadratic equation:

`y=\frac{-(-2)\pm\sqrt[]{(-2)^2-4\cdot1\cdot(-11)_{}}}{2\cdot1}=\frac{2\pm\sqrt[]{48}}{2}=\frac{2\pm4\sqrt[]{3}}{2}=1\pm2\sqrt[]{3}`

The intercepts are (0, 1 - 2sqrt3) and (0, 1 + 2sqrt3)