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
it should read (-3x^2) -4x+1
so how do I write the equation of the parabola in standard form?
thank you
it wants the answer like: y=....(x+-...)^2 +-......
(a) To write the equation of the parabola in standard form, we need to rearrange the given equation to the form y = ax^2 + bx + c, where a, b, and c are constants.
Given y = -3x^2 - 4x + 1, the equation is already in the standard form.
Therefore, the equation of the parabola in standard form is y = -3x^2 - 4x + 1.
(b) To identify the vertex of the parabola, we can use the formula for the x-coordinate of the vertex, given by x = -b / (2a).
In our equation, a = -3 and b = -4. Plugging these values into the formula, we get:
x = -(-4) / (2 * -3)
x = 4 / -6
x = -2 / 3
Now, to find the y-coordinate of the vertex, substitute this x-value back into the equation:
y = -3(-2/3)^2 - 4(-2/3) + 1
y = -3(4/9) + 8/3 + 1
y = -4/3 + 8/3 + 1
y = 5/3
Therefore, the vertex of the parabola is (-2/3, 5/3).
(c) To determine if the vertex is a minimum or maximum, we can look at the coefficient of the x^2 term.
In our equation y = -3x^2 - 4x + 1, the coefficient of the x^2 term is -3, which is negative.
A negative coefficient means that the parabola opens downward, which indicates that the vertex is the maximum point.
Therefore, the vertex of the parabola is a maximum.
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