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 equation \( y = 2x^2 - 5x - 1 \), we need to determine several key features, including the y-intercept, turning point (vertex), and x-intercepts.
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):**
For a quadratic in the form \( y = ax^2 + bx + c \), the x-coordinate of the vertex can be found using the formula \( x = -\frac{b}{2a} \).
Here, \( a = 2 \) and \( b = -5 \):
\[
x = -\frac{-5}{2 \cdot 2} = \frac{5}{4} = 1.25
\]
Now, we can find the corresponding y-coordinate by substituting \( x = 1.25 \) back into the equation:
\[
y = 2(1.25)^2 - 5(1.25) - 1
\]
\[
= 2(1.5625) - 6.25 - 1
\]
\[
= 3.125 - 6.25 - 1
\]
\[
= 3.125 - 7.25 = -4.125
\]
Therefore, the turning point (vertex) is at \( \left(1.25, -4.125\right) \).
3. **X-Intercepts:**
To find the x-intercepts, we set \( y = 0 \):
\[
0 = 2x^2 - 5x - 1
\]
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2 \), \( b = -5 \), and \( c = -1 \):
\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-1)}}{2(2)}
\]
\[
= \frac{5 \pm \sqrt{25 + 8}}{4}
\]
\[
= \frac{5 \pm \sqrt{33}}{4}
\]
So the x-intercepts are:
\[
x_1 = \frac{5 + \sqrt{33}}{4}, \quad x_2 = \frac{5 - \sqrt{33}}{4}
\]
Approximating the values:
\[
\sqrt{33} \approx 5.744
\]
\[
x_1 \approx \frac{5 + 5.744}{4} \approx 2.186, \quad x_2 \approx \frac{5 - 5.744}{4} \approx -0.186
\]
Thus, the x-intercepts are approximately at \( (2.186, 0) \) and \( (-0.186, 0) \).
### Sketch:
1. **Plot points:**
- Y-intercept: \( (0, -1) \)
- Turning point: \( \left(1.25, -4.125\right) \)
- X-intercepts: \( (2.186, 0) \) and \( (-0.186, 0) \)
2. **Shape of the parabola:**
The graph opens upwards since \( a = 2 > 0 \).
3. **Drawing:**
Start at the y-intercept, plot the vertex as the lowest point, and connect these points with a smooth, upward-opening curve, through the x-intercepts.
The final sketch should clearly label these points and indicate the shape of the curve.
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):**
For a quadratic in the form \( y = ax^2 + bx + c \), the x-coordinate of the vertex can be found using the formula \( x = -\frac{b}{2a} \).
Here, \( a = 2 \) and \( b = -5 \):
\[
x = -\frac{-5}{2 \cdot 2} = \frac{5}{4} = 1.25
\]
Now, we can find the corresponding y-coordinate by substituting \( x = 1.25 \) back into the equation:
\[
y = 2(1.25)^2 - 5(1.25) - 1
\]
\[
= 2(1.5625) - 6.25 - 1
\]
\[
= 3.125 - 6.25 - 1
\]
\[
= 3.125 - 7.25 = -4.125
\]
Therefore, the turning point (vertex) is at \( \left(1.25, -4.125\right) \).
3. **X-Intercepts:**
To find the x-intercepts, we set \( y = 0 \):
\[
0 = 2x^2 - 5x - 1
\]
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2 \), \( b = -5 \), and \( c = -1 \):
\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-1)}}{2(2)}
\]
\[
= \frac{5 \pm \sqrt{25 + 8}}{4}
\]
\[
= \frac{5 \pm \sqrt{33}}{4}
\]
So the x-intercepts are:
\[
x_1 = \frac{5 + \sqrt{33}}{4}, \quad x_2 = \frac{5 - \sqrt{33}}{4}
\]
Approximating the values:
\[
\sqrt{33} \approx 5.744
\]
\[
x_1 \approx \frac{5 + 5.744}{4} \approx 2.186, \quad x_2 \approx \frac{5 - 5.744}{4} \approx -0.186
\]
Thus, the x-intercepts are approximately at \( (2.186, 0) \) and \( (-0.186, 0) \).
### Sketch:
1. **Plot points:**
- Y-intercept: \( (0, -1) \)
- Turning point: \( \left(1.25, -4.125\right) \)
- X-intercepts: \( (2.186, 0) \) and \( (-0.186, 0) \)
2. **Shape of the parabola:**
The graph opens upwards since \( a = 2 > 0 \).
3. **Drawing:**
Start at the y-intercept, plot the vertex as the lowest point, and connect these points with a smooth, upward-opening curve, through the x-intercepts.
The final sketch should clearly label these points and indicate the shape of the curve.