Question

graph the quadratic y = 4x^2 + 3x - 8

Answer 1

To graph the quadratic function \( y = 4x^2 + 3x - 8 \), we can follow these steps:

1. **Identify the coefficients**:
- This is in the standard quadratic form \( y = ax^2 + bx + c \), where:
- \( a = 4 \)
- \( b = 3 \)
- \( c = -8 \)

2. **Find the vertex**:
The x-coordinate of the vertex of a parabola described by the equation \( y = ax^2 + bx + c \) can be found using the formula:
\[
x = -\frac{b}{2a}
\]
Substituting our values:
\[
x = -\frac{3}{2 \times 4} = -\frac{3}{8}
\]

Now we can find the corresponding y-coordinate:
\[
y = 4\left(-\frac{3}{8}\right)^2 + 3\left(-\frac{3}{8}\right) - 8
\]
\[
y = 4 \cdot \frac{9}{64} - \frac{9}{8} - 8
\]
\[
y = \frac{36}{64} - \frac{72}{64} - \frac{512}{64}
\]
\[
y = \frac{36 - 72 - 512}{64} = \frac{-548}{64} = -\frac{137}{16}
\]
So the vertex is at \( \left(-\frac{3}{8}, -\frac{137}{16}\right) \).

3. **Find the y-intercept**:
Set \( x = 0 \):
\[
y = 4(0)^2 + 3(0) - 8 = -8
\]
So the y-intercept is \( (0, -8) \).

4. **Find the x-intercepts**:
Set \( y = 0 \):
\[
0 = 4x^2 + 3x - 8
\]
This can be solved using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 4 \cdot -8}}{2 \cdot 4} = \frac{-3 \pm \sqrt{9 + 128}}{8} = \frac{-3 \pm \sqrt{137}}{8}
\]
Thus, the x-intercepts are:
\[
x_1 = \frac{-3 + \sqrt{137}}{8} \quad \text{and} \quad x_2 = \frac{-3 - \sqrt{137}}{8}
\]

5. **Sketch the graph**:
- The vertex \( \left(-\frac{3}{8}, -\frac{137}{16}\right) \) is the lowest point (the parabola opens upwards since \( a = 4 > 0 \)).
- The y-intercept is at \( (0, -8) \).
- The x-intercepts can be approximate; \( \sqrt{137} \) is about \( 11.7 \), so:
- \( x_1 \approx \frac{-3 + 11.7}{8} \approx 1.1 \)
- \( x_2 \approx \frac{-3 - 11.7}{8} \approx -1.8 \)

6. **Plot key points** and draw the parabola:
Start by plotting the vertex, y-intercept, and x-intercepts on a graph. The parabola should be symmetric around the vertex.

Here's a rough sketch based on these calculations:

1. Vertex: \( \left(-\frac{3}{8}, -\frac{137}{16}\right) \approx (-0.375, -8.5625) \)
2. Y-intercept: \( (0, -8) \)
3. X-intercepts: \( \left(\frac{-3 + \sqrt{137}}{8}, 0\right) \) and \( \left(\frac{-3 - \sqrt{137}}{8}, 0\right) \)

This will yield a parabola opening upwards. As I cannot draw directly, you can use graphing tools or graphing paper to plot these points and sketch the parabola accordingly.