Find the coordinates of the vertex and the equation of the axis of symmetry of the parabola given by the equation.
x = − 1-3y^2 + 4
vertex
(x, y) =
axis of symmetry ____
Sketch the graph
I think there's something messed up here.
To find the coordinates of the vertex and the equation of the axis of symmetry of the given parabola, we need to rewrite the equation in the standard form of a parabola, which is y = ax^2 + bx + c.
The given equation is x = -1 - 3y^2 + 4. To transform it into the standard form, we need to isolate the y term on one side of the equation.
x = -1 - 3y^2 + 4
Rearranging the terms, we get:
3y^2 = -x + 3
Divide both sides by 3:
y^2 = (-1/3)x + 1
Now we have the equation in the standard form. Comparing it to the general form y = ax^2 + bx + c, we can see that a = -1/3, b = 0, and c = 1.
Since the coefficient of x is 0, the parabola is symmetric about the y-axis. Therefore, the axis of symmetry is at x = 0.
To find the coordinates of the vertex, we can use the formula x = -b/2a. Since b = 0 in this case, the x-coordinate of the vertex is also 0. To find the y-coordinate, we substitute x = 0 into the equation:
y^2 = (-1/3)(0) + 1
y^2 = 1
Taking the square root of both sides, we get:
y = ±1
So, the coordinates of the vertex are (0, 1) and (0, -1), and the equation of the axis of symmetry is x = 0.
To sketch the graph, plot the vertex at (0, 1) and (0, -1), and draw a parabolic curve that opens either up or down depending on the coefficient of x. In this case, since the coefficient is negative, the parabola opens downwards.