Check the picture below, so the hyperbola looks more or less like so, so let's find the length of the conjugate axis, or namely let's find the "b" component.
![\textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Ctextit%7Bhyperbolas%2C%20horizontal%20traverse%20axis%20%7D%20%5C%5C%5C%5C%20%5Ccfrac%7B%28x-%20h%29%5E2%7D%7B%20a%5E2%7D-%5Ccfrac%7B%28y-%20k%29%5E2%7D%7B%20b%5E2%7D%3D1%20%5Cqquad%20%5Cbegin%7Bcases%7D%20center%5C%20%28%20h%2C%20k%29%5C%5C%20vertices%5C%20%28%20h%5Cpm%20a%2C%20k%29%5C%5C%20c%3D%5Ctextit%7Bdistance%20from%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bcenter%20to%20foci%7D%5C%5C%20%5Cqquad%20%5Csqrt%7B%20a%20%5E2%20%2B%20b%20%5E2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
The equations give you information as to where to plot points.
For y = -x + 1, you know the slope is -1, and the line intersects the y-axis at (0, 1). The y-axis is the vertical line; to plot (0, 1), find 1 on the vertical line and mark it. Now, the slope is -1; that means the line will slope downwards. To plot more points, count 1 unit down from (0, 1) and 1 unit to the right. You should end up at (1, 0).Connect those and you have a line.
For y = -2x + 4, the slope is -2 (so it will also slope downwards), and the y-intercept is 4. Find (0, 4) and plot it. The -2 tells you to count 2 units down (instead of 1 like we did for the last equation) and 1 over. That is the second line.
I hope this helps.
Answer: C add 4 to both sides
I hope this helps you !
Answer:
haha
Step-by-step explanation:
