Nitha Mulungu
Aug 14
Sketch the graph of equation y=-3x^2 +2x -1 ?
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ChatGPT
Aug 14
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.
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Google Gemini AI
Aug 14
Step 1: Find the vertex.
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).
Step 2: Find the y-intercept.
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).
Step 3: Find the x-intercepts.
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).
Step 4: Sketch the graph.
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)]
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).
Step 2: Find the y-intercept.
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).
Step 3: Find the x-intercepts.
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).
Step 4: Sketch the graph.
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)]
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