y=−3x2−4x+1
(a) Write the equation of the parabola in standard form
(b) Identify the vertex of the parabola
C)Is the vertex a minimum or a maximum?
many thanks
4 answers
it should read (-3x^2) -4x+1
y = -3x^2 - 4 x + 1
convert to vertex form
3 x^2 + 4 x = - y + 1
x^2 + (4/3) x = - y/3 + 1/3
x^2 + (4/3) x + 4/9 = -y/3 +7/9
(x+2/3)^2 = -(1/3)(y - 7/3)
vertex at x = -2/3 , y = 7/3
as x gets big positive or negative, y goes down, sheds water, opens down, vertex at the top
convert to vertex form
3 x^2 + 4 x = - y + 1
x^2 + (4/3) x = - y/3 + 1/3
x^2 + (4/3) x + 4/9 = -y/3 +7/9
(x+2/3)^2 = -(1/3)(y - 7/3)
vertex at x = -2/3 , y = 7/3
as x gets big positive or negative, y goes down, sheds water, opens down, vertex at the top
so how do I write the equation of the parabola in standard form?
thank you
thank you
it wants the answer like: y=....(x+-...)^2 +-......
1 answer