How to graph -3x^2+12x-4?

Question

kay so I have the y intercept as -4. I used the formula -12/2(-3) to get 2 and substituted it into the equation and got 8. So (2,8) is the vertex. I did 3x^2-6x-2x+4 to get (3x-2)(x-2). I set them both equal to 0 and got 2/3 & 2 for the zeroes. But when I graph that I dont get a parabola. Can someone please tell me what I did wrong?

Answer 1

The function f (x) = y = - 3x² + 12x - 4.

The above function represents a parabola.

Graph of the parabola in vertex form :

The vertex form of parabola equation is y = a(x - h)^2 + k, where (h, k) = vertex and axis of symmetry x = h.

The parabola is f(x) = y = - 3x² + 12x - 4.

Write the equation in vertex form of a parabola equation.

Divide each side by negative 3.

- y/3 = x² - 4x + 4/3

To change the expression [x² - 4x + 4/3] into a perfect square trinomial add and subtract (half the x coefficient)²

 Here x coefficient = - 4. so, (half the x coefficient)² = (- 4/2)²= 4.

- y/3 = x² - 4x + 4 + 4/3 - 4

- y/3 = (x - 2)² + (4 - 12)/3

- y/3 = (x - 2)² - 8/3

y = - 3(x - 2)² + 8.

Compare the equation above equation with the vertex form of parabola equation.

Vertex (h, k) = (2, 8), and axis of symmetry x = 2.

Make the table of values to find ordered pairs that satisfy the equation.

Choose values for x and find the corresponding values for y.

x	y = - 3(x - 2)2 + 8	(x, y)
0	y = - 3(0 - 2)2 + 8 = - 3(- 2)2 + 8 = - 12 + 8 = - 4	(0, - 4)
1	y = - 3(1 - 2)2 + 8 = - 3(- 1)2 + 8 = - 3 + 8 = 5	(1, 5)
2	y = - 3(2 - 2)2 + 8 = 0 + 8 = 8	(2, 8)
3	y = - 3(3 - 2)2 + 8 = - 3(1)2 + 8 = - 3 + 8 = 5	(3, 5)
4	y = - 3(4 - 2)2 + 8 = - 3(2)2 + 8 = - 12 + 8 = - 4	(4, - 4)

1.Draw a coordinate plane.

2.Plot the coordinate points.

3.Then sketch the graph, connecting the points with a smooth curve.

Graph :