Question

Find the coordinates lf the parabola y = 5x^2 + 8x + 9

Answer 1

To find the coordinates of the parabola given by the equation \( y = 5x^2 + 8x + 9 \), we can identify several key features, including the vertex, x-intercepts, and y-intercept.

1. **Vertex**: The vertex of a parabola represented by the equation \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex:
\[
x = -\frac{b}{2a}
\]
In this case, \( a = 5 \) and \( b = 8 \):
\[
x = -\frac{8}{2 \cdot 5} = -\frac{8}{10} = -\frac{4}{5}
\]
Now, we substitute \( x = -\frac{4}{5} \) back into the equation to find the y-coordinate of the vertex:
\[
y = 5\left(-\frac{4}{5}\right)^2 + 8\left(-\frac{4}{5}\right) + 9
\]
\[
= 5\left(\frac{16}{25}\right) - \frac{32}{5} + 9
\]
\[
= \frac{80}{25} - \frac{160}{25} + \frac{225}{25}
\]
\[
= \frac{80 - 160 + 225}{25} = \frac{145}{25} = \frac{29}{5}
\]
Therefore, the vertex of the parabola is at:
\[
\left(-\frac{4}{5}, \frac{29}{5}\right)
\]

2. **Y-intercept**: The y-intercept can be found by substituting \( x = 0 \) into the equation:
\[
y = 5(0)^2 + 8(0) + 9 = 9
\]
So, the y-intercept is at the point \( (0, 9) \).

3. **X-intercepts**: To find the x-intercepts, we set \( y = 0 \):
\[
0 = 5x^2 + 8x + 9
\]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 5 \cdot 9}}{2 \cdot 5}
\]
\[
= \frac{-8 \pm \sqrt{64 - 180}}{10}
\]
\[
= \frac{-8 \pm \sqrt{-116}}{10}
\]
Since the discriminant is negative, there are no real x-intercepts, indicating that the parabola does not intersect the x-axis.

### Summary of Coordinates
- **Vertex**: \( \left(-\frac{4}{5}, \frac{29}{5}\right) \)
- **Y-intercept**: \( (0, 9) \)
- **X-intercepts**: None (no real solutions)

If you have any further calculations or specific points to analyze, let me know!