Question
sketch the graph of y=2x^2-5x+1,showing clearly the y-intercept ,turning point and the x-intercepts?
Answers
GPT-4o mini
To sketch the graph of the quadratic function \(y = 2x^2 - 5x + 1\), we will identify its key features: the y-intercept, the turning point (vertex), and the x-intercepts (roots).
### 1. Y-intercept:
The y-intercept occurs when \(x = 0\).
\[
y = 2(0)^2 - 5(0) + 1 = 1
\]
So, the y-intercept is at the point \((0, 1)\).
### 2. Turning Point (Vertex):
To find the vertex of the parabola, we can use the formula for the x-coordinate of the vertex, given by:
\[
x = -\frac{b}{2a}
\]
where \(a = 2\) and \(b = -5\).
\[
x = -\frac{-5}{2(2)} = \frac{5}{4} = 1.25
\]
Now, substitute \(x = 1.25\) back into the original equation to find the y-coordinate of the vertex:
\[
y = 2(1.25)^2 - 5(1.25) + 1
\]
\[
y = 2(1.5625) - 6.25 + 1
\]
\[
y = 3.125 - 6.25 + 1 = -2.125
\]
Thus, the turning point (vertex) of the parabola is at \(\left(1.25, -2.125\right)\).
### 3. X-intercepts:
To find the x-intercepts, we need to solve for \(x\) when \(y = 0\):
\[
2x^2 - 5x + 1 = 0
\]
We can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For our coefficients, \(a = 2\), \(b = -5\), and \(c = 1\):
\[
x = \frac{5 \pm \sqrt{(-5)^2 - 4(2)(1)}}{2(2)}
\]
\[
= \frac{5 \pm \sqrt{25 - 8}}{4}
\]
\[
= \frac{5 \pm \sqrt{17}}{4}
\]
Calculating this gives us two x-intercepts:
\[
x_1 = \frac{5 + \sqrt{17}}{4} \quad \text{and} \quad x_2 = \frac{5 - \sqrt{17}}{4}
\]
### Summary of Key Points:
- Y-intercept: \((0, 1)\)
- Turning point (Vertex): \(\left(1.25, -2.125\right)\)
- X-intercepts:
- \(x_1 \approx 3.28\) (exactly \(\frac{5 + \sqrt{17}}{4}\))
- \(x_2 \approx 0.22\) (exactly \(\frac{5 - \sqrt{17}}{4}\))
### Sketch of the Graph:
To sketch the graph, plot the points:
1. The y-intercept at \((0, 1)\)
2. The vertex at \(\left(1.25, -2.125\right)\)
3. The x-intercepts at \(\left(\frac{5 - \sqrt{17}}{4}, 0\right)\) and \(\left(\frac{5 + \sqrt{17}}{4}, 0\right)\)
The parabola opens upwards (since the coefficient of \(x^2\) is positive), and the turning point is the lowest point on the graph. Connect these points smoothly to complete the sketch.
Here’s a visualization of what the graph would generally look like (without drawing software):
```
|
3 | *
|
2 | *
|
1 | * *
|
+---------*---------*---------*---------
-2.125 | *
|
-1 |
|
-2 |
|
-3 |
|
|
-4 |
|
-5 |
```
**Note:** The exact placement of x-intercepts and the scaling of axes would depend on a complete graphing environment. The key points should be accurately given on a graph for precise representation.
### 1. Y-intercept:
The y-intercept occurs when \(x = 0\).
\[
y = 2(0)^2 - 5(0) + 1 = 1
\]
So, the y-intercept is at the point \((0, 1)\).
### 2. Turning Point (Vertex):
To find the vertex of the parabola, we can use the formula for the x-coordinate of the vertex, given by:
\[
x = -\frac{b}{2a}
\]
where \(a = 2\) and \(b = -5\).
\[
x = -\frac{-5}{2(2)} = \frac{5}{4} = 1.25
\]
Now, substitute \(x = 1.25\) back into the original equation to find the y-coordinate of the vertex:
\[
y = 2(1.25)^2 - 5(1.25) + 1
\]
\[
y = 2(1.5625) - 6.25 + 1
\]
\[
y = 3.125 - 6.25 + 1 = -2.125
\]
Thus, the turning point (vertex) of the parabola is at \(\left(1.25, -2.125\right)\).
### 3. X-intercepts:
To find the x-intercepts, we need to solve for \(x\) when \(y = 0\):
\[
2x^2 - 5x + 1 = 0
\]
We can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For our coefficients, \(a = 2\), \(b = -5\), and \(c = 1\):
\[
x = \frac{5 \pm \sqrt{(-5)^2 - 4(2)(1)}}{2(2)}
\]
\[
= \frac{5 \pm \sqrt{25 - 8}}{4}
\]
\[
= \frac{5 \pm \sqrt{17}}{4}
\]
Calculating this gives us two x-intercepts:
\[
x_1 = \frac{5 + \sqrt{17}}{4} \quad \text{and} \quad x_2 = \frac{5 - \sqrt{17}}{4}
\]
### Summary of Key Points:
- Y-intercept: \((0, 1)\)
- Turning point (Vertex): \(\left(1.25, -2.125\right)\)
- X-intercepts:
- \(x_1 \approx 3.28\) (exactly \(\frac{5 + \sqrt{17}}{4}\))
- \(x_2 \approx 0.22\) (exactly \(\frac{5 - \sqrt{17}}{4}\))
### Sketch of the Graph:
To sketch the graph, plot the points:
1. The y-intercept at \((0, 1)\)
2. The vertex at \(\left(1.25, -2.125\right)\)
3. The x-intercepts at \(\left(\frac{5 - \sqrt{17}}{4}, 0\right)\) and \(\left(\frac{5 + \sqrt{17}}{4}, 0\right)\)
The parabola opens upwards (since the coefficient of \(x^2\) is positive), and the turning point is the lowest point on the graph. Connect these points smoothly to complete the sketch.
Here’s a visualization of what the graph would generally look like (without drawing software):
```
|
3 | *
|
2 | *
|
1 | * *
|
+---------*---------*---------*---------
-2.125 | *
|
-1 |
|
-2 |
|
-3 |
|
|
-4 |
|
-5 |
```
**Note:** The exact placement of x-intercepts and the scaling of axes would depend on a complete graphing environment. The key points should be accurately given on a graph for precise representation.