To find the vertex of the quadratic equation \( y = 2x^2 + 5x - 10 \), we can use the formula for the vertex of a parabola given by the equation \( y = ax^2 + bx + c \). The x-coordinate of the vertex can be found using the formula:
\[ x = -\frac{b}{2a} \]
Here, \( a = 2 \) and \( b = 5 \).
- Calculate the x-coordinate of the vertex:
\[ x = -\frac{5}{2 \cdot 2} = -\frac{5}{4} = -1.25 \]
- Calculate the y-coordinate of the vertex by substituting \( x = -1.25 \) back into the quadratic equation:
\[ y = 2(-1.25)^2 + 5(-1.25) - 10 \]
Calculating \( (-1.25)^2 \):
\[ (-1.25)^2 = 1.5625 \]
Substituting this value back into the equation:
\[ y = 2(1.5625) + 5(-1.25) - 10 \]
Now calculate:
\[ y = 3.125 - 6.25 - 10 \] \[ y = 3.125 - 16.25 = -13.125 \]
Thus, the vertex of the quadratic \( y = 2x^2 + 5x - 10 \) is:
\[ (-1.25, -13.125) \]
In rounded form (to the nearest hundredth, if needed):
- Vertex: (-1.25, -13.13)
Graphing the Quadratic
To graph the quadratic, we can plot the vertex and a few additional points. Here's a simple way to do so:
- Vertex: \( (-1.25, -13.13) \)
- Choose x-values around the vertex, for example, \( x = -3, -2, -1, 0, 1 \).
Calculating y for these:
-
\( x = -3 \): \[ y = 2(-3)^2 + 5(-3) - 10 = 18 - 15 - 10 = -7 \]
-
\( x = -2 \): \[ y = 2(-2)^2 + 5(-2) - 10 = 8 - 10 - 10 = -12 \]
-
\( x = -1 \): \[ y = 2(-1)^2 + 5(-1) - 10 = 2 - 5 - 10 = -13 \]
-
\( x = 0 \): \[ y = 2(0)^2 + 5(0) - 10 = -10 \]
-
\( x = 1 \): \[ y = 2(1)^2 + 5(1) - 10 = 2 + 5 - 10 = -3 \]
Now we can plot the points:
- \( (-3, -7) \)
- \( (-2, -12) \)
- \( (-1, -13) \)
- \( (0, -10) \)
- \( (1, -3) \)
This will help us sketch the parabola.
The parabola opens upwards (since \( a = 2 > 0 \)), and the graph will be symmetric about the vertical line through the vertex.
Summary
- Vertex: \((-1.25, -13.13)\)