y=5x^2-10
The given equation is a quadratic function in the form of \( y = ax^2 + bx + c \), where:
- \( a = 5 \)
- \( b = 0 \)
- \( c = -10 \)
Characteristics of the Quadratic Function
Vertex: The vertex of a parabola given by \( y = ax^2 + bx + c \) can be found using the formula:
\[ x = -\frac{b}{2a} \]
Plugging in the values: \[ x = -\frac{0}{2 \cdot 5} = 0 \]
Now, substitute \( x = 0 \) back into the original equation to find the \( y \)-coordinate: \[ y = 5(0)^2 - 10 = -10 \]
So, the vertex of the parabola is \( (0, -10) \).
Direction: Since \( a = 5 > 0 \), the parabola opens upwards.
Axis of Symmetry: The axis of symmetry is the vertical line that passes through the vertex. In this case, it is \( x = 0 \).
Y-Intercept: To find the y-intercept, set \( x = 0 \): \[ y = 5(0)^2 - 10 = -10 \] Thus, the y-intercept is \( (0, -10) \).
X-Intercepts: To find the x-intercepts, set \( y = 0 \): \[ 0 = 5x^2 - 10 \] Solving for \( x \): \[ 5x^2 = 10 \\ x^2 = 2 \\ x = \pm\sqrt{2} \]
Therefore, the x-intercepts are \( (\sqrt{2}, 0) \) and \( (-\sqrt{2}, 0) \).
Summary
- Vertex: \( (0, -10) \)
- Direction: Opens upwards
- Axis of Symmetry: \( x = 0 \)
- Y-Intercept: \( (0, -10) \)
- X-Intercepts: \( (\sqrt{2}, 0) \) and \( (-\sqrt{2}, 0) \)
If you need any further calculations or clarifications, feel free to ask!
