Consider the parabola given by the equation: f(x)=-2x^2+6x+8


Find the following for this parabola:

To analyze the parabola given by the equation \( f(x) = -2x^2 + 6x + 8 \), we can find the following characteristics:

1. **Vertex**: The vertex of a parabola in the form \( f(x) = ax^2 + bx + c \) can be found using the formula:
   \[
   x = -\frac{b}{2a}
   \]
   Here, \( a = -2 \) and \( b = 6 \):
   \[
   x = -\frac{6}{2 \cdot -2} = \frac{6}{4} = \frac{3}{2}
   \]
   To find the \( y \)-coordinate of the vertex, substitute \( x = \frac{3}{2} \) back into the function:
   \[
   f\left(\frac{3}{2}\right) = -2\left(\frac{3}{2}\right)^2 + 6\left(\frac{3}{2}\right) + 8
   \]
   \[
   = -2 \cdot \frac{9}{4} + 9 + 8 = -\frac{18}{4} + 9 + 8 = -\frac{9}{2} + 9 + 8
   \]
   \[
   = -\frac{9}{2} + \frac{18}{2} + \frac{16}{2} = \frac{25}{2}
   \]
   So, the vertex is at \( \left(\frac{3}{2}, \frac{25}{2}\right) \).

2. **Axis of Symmetry**: The axis of symmetry is a vertical line that passes through the vertex. Thus, the equation of the axis of symmetry is:
   \[
   x = \frac{3}{2}
   \]

3. **Y-Intercept**: The \( y \)-intercept can be found by evaluating \( f(0) \):
   \[
   f(0) = -2(0)^2 + 6(0) + 8 = 8
   \]
   Thus, the \( y \)-intercept is \( (0, 8) \).

4. **X-Intercepts**: To find the \( x \)-intercepts, set \( f(x) = 0 \):
   \[
   -2x^2 + 6x + 8 = 0
   \]
   Rearranging gives:
   \[
   2x^2 - 6x - 8 = 0
   \]
   Dividing through by 2:
   \[
   x^2 - 3x - 4 = 0
   \]
   Factoring gives:
   \[
   (x - 4)(x + 1) = 0
   \]
   Thus, \( x = 4 \) and \( x = -1 \). The \( x \)-intercepts are \( (4, 0) \) and \( (-1, 0) \).

5. **Direction**: Since the coefficient of \( x^2 \) (which is \( -2 \)) is negative, the parabola opens downwards.

### Summary of Results:
- Vertex: \( \left(\frac{3}{2}, \frac{25}{2}\right) \)
- Axis of Symmetry: \( x = \frac{3}{2} \)
- Y-Intercept: \( (0, 8) \)
- X-Intercepts: \( (4, 0) \) and \( (-1, 0) \)
- Direction: Opens downwards

Would you like more details or do you have any questions? Here are some related questions to consider:

1. What are the coordinates of the focus of this parabola?
2. How do you determine the length of the latus rectum for this parabola?
3. What would the vertex be if the parabola were expressed in vertex form?
4. How can you graph this parabola accurately?
5. What are the effects of changing the coefficients of the quadratic equation on the graph? 

**Tip**: When finding the vertex, always check if the parabola opens upwards or downwards to understand its orientation.

Consider the parabola given by the equation: f(x)=-2x^2+6x+8. Find the vertex, axis of symmetry, y-intercept, x-intercepts, and direction.

Explore the analysis of the parabola defined by f(x)=-2x^2+6x+8. Learn how to find its vertex at (3/2, 25/2), axis of symmetry x=3/2, y-intercept (0, 8), and x-intercepts (4, 0) and (-1, 0). This guide is suitable for students in Grades 8-10.

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