y=-3x^2+6x-2

Sketch the graph of the given porabola and state the coordinance of its intercepts.
I can graph it if I can get some help with the vertex and coordinance.
This is what I got; vertex:(1,1) x-intercept:(-1.6,.-4), and y-intercept:(0,-2). Is this right?

that's suppose to be -.4

what is thr perimeter of an equilateral triangle that measurea 4 in one side

Rachel: 12. Perimeter means to "walk around" the entire object. If the side of an equilateral triangle is 4, so are the other 2 sides (4 + 4 + 4).

No, Jennifer, you are close.

How can you have an x-intercept of (-1.6, -.4) ??
x-intercepts have the form (?,0)

Your vertex is correct, and your y-intercept is correct

to find the x-intercepts, let y = 0
0 = -3x^2+6x-2
3x^2 - 6x + 2 = 0
x = (6 ± √12)/6
= 1.577 or .423

those are the x values of the intercepts
so the x-intercepts are
(1.6,0) and (.4,0)

(you actually had the opposite of these numbers, so you must have made a sign error.
Also notice that for any parabola the midpoint of the x-intercepts gives you the x value of the vertex. That would not happen for your values, but
(1.577 + .423)/2 = 1, the x value of the vertex)

To sketch the graph of the given quadratic equation y = -3x^2 + 6x - 2, you can follow these steps:

1. Determine the vertex of the parabola:
The x-coordinate of the vertex can be found using the formula: x = -b/2a, where a, b, and c are the coefficients of the quadratic equation in the form of ax^2 + bx + c.
In this case, a = -3 and b = 6, so the x-coordinate of the vertex is x = -6 / (2 * -3) = -6 / -6 = 1.
To find the y-coordinate of the vertex, substitute this x-value into the equation: y = -3(1)^2 + 6(1) - 2 = -3 + 6 - 2 = 1.
Therefore, the vertex of the parabola is (1, 1).

2. Locate the x-intercepts:
To find the x-intercepts, set y = 0 in the quadratic equation and solve for x.
0 = -3x^2 + 6x - 2
This equation cannot be factored easily, so you can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a.
In this case, a = -3, b = 6, and c = -2. Substituting these values into the formula, we get:
x = (-6 ± √(6^2 - 4(-3)(-2))) / (2 * -3)
Simplifying further, we get:
x = (-6 ± √(36 - 24)) / (-6)
x = (-6 ± √12) / (-6)
x = (-6 ± 2√3) / (-6)
Simplifying the expression, we have:
x = 1 ± √3/3
So the two x-intercepts are approximately:
x-intercept 1: (-1.54, 0)
x-intercept 2: (2.21, 0)

3. Find the y-intercept:
To find the y-intercept, set x = 0 in the quadratic equation and solve for y.
y = -3(0)^2 + 6(0) - 2
y = 0 - 0 - 2
y = -2
Therefore, the y-intercept is (0, -2).

Based on these calculations, the sketch of the parabola should have the following key points:
- Vertex: (1, 1)
- X-intercepts: (-1.54, 0) and (2.21, 0)
- Y-intercept: (0, -2)

Please note that these values are approximate as they have been rounded to two decimal places.