Saved Bookmarks
| 1. |
Find the value of `P`such that the vertex of `y=x^2+2p x+13`is 4 units above the x-axis. |
|
Answer» The given parabola is `y=x^(2)+2px+13` `ory=(x+p)^(2)+13-p^(2)` `ory-(13-p^(2))=(x+p)^(2)` So, the vertex is `(-p,13-p^(2))`. Since the parabola is at a distance of 4 units above the x-axis, `13-p^(2)=4` `orp^(2)=9` `orp=pm3` |
|