Find the vertex and axis of symmetry of the parabola when y=-3x^2+12-8
Don't you mean
y = 3x^2 +12x -8 ?
If so, "complete the square" and rewrite it as
y = 3(x^2 + 4x + 4)- 20
= 3(x+2)^2 -20
That is symmetric about x = -2, and has a vertex at
x = -2, y = -20
how to graph
To find the vertex and axis of symmetry of a parabola in the form of y = ax^2 + bx + c, we can use the equation:
x = -b / (2a)
In the given parabola, y = -3x^2 + 12x - 8, we can determine the vertex and axis of symmetry by following these steps:
Step 1: Identify the coefficients
In our equation, a = -3, b = 12, and c = -8.
Step 2: Calculate x-coordinate of the vertex
Using the equation x = -b / (2a), we substitute the values:
x = -12 / (2 * -3)
x = -12 / -6
x = 2
Step 3: Substitute x-coordinate into the original equation to find the y-coordinate
Using the x-coordinate we found, let's substitute it back into the original equation:
y = -3(2)^2 + 12(2) - 8
y = -12 + 24 - 8
y = 4
So, the vertex of the parabola is located at the point (2, 4).
Step 4: Determine the axis of symmetry
The axis of symmetry is a vertical line that passes through the vertex. Since the x-coordinate of the vertex is 2, the equation of the axis of symmetry is x = 2.
Therefore, the vertex of the parabola is (2, 4) and the axis of symmetry is x = 2.