y^2+5y-x-6=0?

Question

answer (a)determine the vertex of the parabola
(b)what are the x-intercepts and y-intercepts
(c) sketch the parabola

Answer 1

well, in this case, you have to complete the square in the y side..isolating y terms on left.. you end with

y^2 +5y = x+6

completing square on left side, gives us..

y^2 +5y +(5/2)^2= x+6+25/4

factoring left side..

(y+5/2)^2 = x+49/4

vertex then, is at (-49/4, -5/2)

letting x=0, gives us the y intercepts, which happen at
y= -5/2 +- 7/2 or.. -6, and 1

if on the other hand, letting y=0

gives us x=-5

Answer 2

get vertex by rewriting in std form:

y² + 5y = x + 6
y² + 5y + 25/4 = x + 6 + 25/4
(y + 5/2)² = x + 49/4
so the vertex is (-12.25, -2.5)

x intercepts when y = 0:
-x - 6 = 0
x = -6, so (-6,0)

y intercepts when x = 0
y² + 5y - 6 = 0
(y+6)(y-1) = 0
y = -6, y = 1, so (0,-6) and (0,1)

sketch by drawing the parabola through the known points, given that it opens to the right.

Answer 3

yes

Answer 4

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