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more than 4 weeks ago

To sketch the graph of the equation y = -3x^2 + 2x - 1, we can start by finding the vertex of the parabola. The vertex of a parabola in the form y = ax^2 + bx + c is given by the formula x = -b/2a.

In this case, a = -3 and b = 2. So the x-coordinate of the vertex is x = -2/(2*(-3)) = 1/3.

To find the y-coordinate of the vertex, we can substitute x = 1/3 into the equation y = -3x^2 + 2x - 1:

y = -3(1/3)^2 + 2(1/3) - 1

y = -3(1/9) + 2/3 - 1

y = -1/3 + 2/3 - 1

y = -1/3

So the vertex of the parabola is at (1/3, -1/3).

Next, we can find the x-intercepts by setting y = 0 and solving for x:

0 = -3x^2 + 2x - 1

3x^2 - 2x + 1 = 0

This quadratic equation does not factor easily, so we can use the quadratic formula to find the x-intercepts:

x = (-(-2) ± sqrt((-2)^2 - 4*3*1))/(2*3)

x = (2 ± sqrt(4 - 12))/6

x = (2 ± sqrt(-8))/6

x = (2 ± 2i√2)/6

x = (1 ± i√2)/3

So the x-intercepts are at (1/3 + i√2/3, 0) and (1/3 - i√2/3, 0).

Now we can sketch the graph of the equation y = -3x^2 + 2x - 1. The parabola opens downwards because the coefficient of x^2 is negative. The vertex is at (1/3, -1/3) and the x-intercepts are at (1/3 + i√2/3, 0) and (1/3 - i√2/3, 0). The graph should look like a downward-facing parabola passing through these points.

In this case, a = -3 and b = 2. So the x-coordinate of the vertex is x = -2/(2*(-3)) = 1/3.

To find the y-coordinate of the vertex, we can substitute x = 1/3 into the equation y = -3x^2 + 2x - 1:

y = -3(1/3)^2 + 2(1/3) - 1

y = -3(1/9) + 2/3 - 1

y = -1/3 + 2/3 - 1

y = -1/3

So the vertex of the parabola is at (1/3, -1/3).

Next, we can find the x-intercepts by setting y = 0 and solving for x:

0 = -3x^2 + 2x - 1

3x^2 - 2x + 1 = 0

This quadratic equation does not factor easily, so we can use the quadratic formula to find the x-intercepts:

x = (-(-2) ± sqrt((-2)^2 - 4*3*1))/(2*3)

x = (2 ± sqrt(4 - 12))/6

x = (2 ± sqrt(-8))/6

x = (2 ± 2i√2)/6

x = (1 ± i√2)/3

So the x-intercepts are at (1/3 + i√2/3, 0) and (1/3 - i√2/3, 0).

Now we can sketch the graph of the equation y = -3x^2 + 2x - 1. The parabola opens downwards because the coefficient of x^2 is negative. The vertex is at (1/3, -1/3) and the x-intercepts are at (1/3 + i√2/3, 0) and (1/3 - i√2/3, 0). The graph should look like a downward-facing parabola passing through these points.

Google Gemini AI

more than 4 weeks ago

The vertex of a parabola is given by the formula (-b/2a, f(-b/2a)). In this case, a = -3 and b = 2, so the x-coordinate of the vertex is:

```

-b/2a = -2/2(-3) = 1/3

```

To find the y-coordinate, we evaluate the function at x = 1/3:

```

f(-b/2a) = f(1/3) = -3(1/3)^2 + 2(1/3) - 1 = -1/3

```

Therefore, the vertex is located at (1/3, -1/3).

The y-intercept is the point where the parabola crosses the y-axis. To find it, we set x = 0:

```

y = -3(0)^2 + 2(0) - 1 = -1

```

Therefore, the y-intercept is (0, -1).

The x-intercepts are the points where the parabola crosses the x-axis. To find them, we set y = 0:

```

0 = -3x^2 + 2x - 1

3x^2 - 2x + 1 = 0

(3x - 1)(x - 1) = 0

x = 1/3 or x = 1

```

Therefore, the x-intercepts are (1/3, 0) and (1, 0).

Using the information we have gathered, we can sketch the graph of the parabola as follows:

[Image of a parabola that opens downwards, with vertex at (1/3, -1/3), y-intercept at (0, -1), and x-intercepts at (1/3, 0) and (1, 0)]