To analyze the quadratic expression $3x^2 + 5x + 2$, we can break it down into several steps. Here’s a detailed analysis of the expression:

1. Standard Form of a Quadratic

The given expression is already in standard quadratic form: $f(x) = ax^2 + bx + c$ where:

• $a = 3$
• $b = 5$
• $c = 2$

2. Vertex Form

To find the vertex of the parabola represented by this quadratic, we can complete the square or use the vertex formula. The vertex $x$-coordinate is given by: $x_{\text{vertex}} = \frac{-b}{2a}$ Substitute $a = 3$ and $b = 5$: $x_{\text{vertex}} = \frac{-5}{2(3)} = \frac{-5}{6}$ Now substitute this back into the original quadratic equation to find the $y$-coordinate of the vertex: $y_{\text{vertex}} = 3\left(\frac{-5}{6}\right)^2 + 5\left(\frac{-5}{6}\right) + 2$ This simplifies to: $y_{\text{vertex}} = 3\left(\frac{25}{36}\right) - \frac{25}{6} + 2$ $y_{\text{vertex}} = \frac{75}{36} - \frac{150}{36} + \frac{72}{36}$ $y_{\text{vertex}} = \frac{-3}{36} = \frac{-1}{12}$ So the vertex is at $\left( \frac{-5}{6}, \frac{-1}{12} \right)$.

3. Discriminant

The discriminant of a quadratic equation is used to determine the nature of the roots. It’s given by the formula: $\Delta = b^2 - 4ac$ Substitute $a = 3$, $b = 5$, and $c = 2$: $\Delta = 5^2 - 4(3)(2) = 25 - 24 = 1$ Since the discriminant is positive ($\Delta = 1$), this means the quadratic has two distinct real roots.

4. Roots of the Quadratic

The roots can be found using the quadratic formula: $x = \frac{-b \pm \sqrt{\Delta}}{2a}$ Substitute $b = 5$, $\Delta = 1$, and $a = 3$: $x = \frac{-5 \pm \sqrt{1}}{2(3)} = \frac{-5 \pm 1}{6}$ Thus, the two roots are: $x_1 = \frac{-5 + 1}{6} = \frac{-4}{6} = \frac{-2}{3}$ $x_2 = \frac{-5 - 1}{6} = \frac{-6}{6} = -1$ So the roots are $x = -\frac{2}{3}$ and $x = -1$.

5. Graph Behavior

• Since $a = 3 > 0$, the parabola opens upward.
• The vertex $\left( \frac{-5}{6}, \frac{-1}{12} \right)$ is the minimum point of the parabola.
• The roots of the quadratic, $x = -\frac{2}{3}$ and $x = -1$, are where the graph crosses the x-axis.

6. Intercepts

• Y-intercept: Set $x = 0$ in the original equation: $y = 3(0)^2 + 5(0) + 2 = 2$ So, the y-intercept is $(0, 2)$.

• X-intercepts: Already found as the roots: $x = -\frac{2}{3}$ and $x = -1$.

Summary of the Analysis:

• Vertex: $\left( \frac{-5}{6}, \frac{-1}{12} \right)$
• Discriminant: $\Delta = 1$ (two real roots)
• Roots: $x = -\frac{2}{3}$ and $x = -1$
• Parabola opens upwards (since $a = 3 > 0$)
• Y-intercept: $(0, 2)$

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Would you like a more detailed breakdown on any of the steps above? Or do you have other questions related to quadratic functions?

5 Follow-up Questions:

1. How can we derive the vertex form of this quadratic equation?
2. How does the discriminant affect the number of real solutions in a quadratic equation?
3. What happens if the discriminant is negative?
4. How can you graph this quadratic equation manually on graph paper?
5. How would this analysis change if the quadratic equation had a negative $a$-value (i.e., if it opened downward)?

Tip: The vertex of a parabola provides key information: it gives the minimum or maximum value of the quadratic function, which can be very helpful when solving optimization problems!