Let's analyze the function $y = 3x^2 + 2x + 4$ and graph it without using a calculator.

1. Identify the Type of Function:

• The function $y = 3x^2 + 2x + 4$ is a quadratic function, and its graph will be a parabola. Since the coefficient of $x^2$ is positive (3), the parabola opens upwards.

2. Find the Vertex:

• The vertex form of a quadratic function is $y = a(x-h)^2 + k$, where $(h, k)$ is the vertex of the parabola.
• To find the vertex for $y = 3x^2 + 2x + 4$, we use the vertex formula: $h = -\frac{b}{2a} = -\frac{2}{2 \times 3} = -\frac{2}{6} = -\frac{1}{3}$
• Substitute $h = -\frac{1}{3}$ back into the equation to find $k$: $k = 3\left(-\frac{1}{3}\right)^2 + 2\left(-\frac{1}{3}\right) + 4$ $k = 3\left(\frac{1}{9}\right) - \frac{2}{3} + 4 = \frac{1}{3} - \frac{2}{3} + 4 = \frac{-1}{3} + 4 = \frac{11}{3}$
• The vertex is $\left(-\frac{1}{3}, \frac{11}{3}\right)$.

3. Determine the Axis of Symmetry:

• The axis of symmetry is the vertical line through the vertex: $x = -\frac{1}{3}$

4. Find the Y-Intercept:

• The y-intercept occurs when $x = 0$: $y = 3(0)^2 + 2(0) + 4 = 4$
• The y-intercept is $(0, 4)$.

5. Find the X-Intercepts:

• The x-intercepts occur where $y = 0$: $3x^2 + 2x + 4 = 0$
• Solving this quadratic equation using the discriminant: $\text{Discriminant} = b^2 - 4ac = 2^2 - 4(3)(4) = 4 - 48 = -44$
• Since the discriminant is negative, there are no real x-intercepts. The parabola does not cross the x-axis.

6. Determine the Asymptotic Behavior:

• Since this is a polynomial function (quadratic), it has no asymptotes. As $x$ approaches infinity or negative infinity, the function will continue to increase due to the positive leading coefficient.

7. Sketch the Graph:

• Plot the vertex $\left(-\frac{1}{3}, \frac{11}{3}\right)$.
• Draw the axis of symmetry $x = -\frac{1}{3}$.
• Plot the y-intercept $(0, 4)$.
• Since the parabola opens upwards and does not have real x-intercepts, sketch the parabola passing through these points with the correct curvature.

8. Summary of Features:

• Vertex: $\left(-\frac{1}{3}, \frac{11}{3}\right)$
• Axis of Symmetry: $x = -\frac{1}{3}$
• Y-Intercept: $(0, 4)$
• X-Intercepts: None (no real solutions)
• Opening: Upward
• Asymptotes: None

Would you like a more detailed explanation of any part of this process or have any questions?

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Here are some related questions you might consider:

1. How would the graph change if the coefficient of $x^2$ were negative?
2. What is the effect of changing the linear coefficient $b$ on the vertex of the parabola?
3. How do you determine the direction of opening for a quadratic function?
4. What is the significance of the discriminant in finding the x-intercepts?
5. How would you graph $y = 3x^2 + 2x + 4$ using transformations from the parent function $y = x^2$?
6. How can you convert the given quadratic equation into vertex form?
7. What is the role of the constant term in a quadratic function?
8. How do you determine the concavity of a parabola from the quadratic function?

Tip: The discriminant in a quadratic equation provides key information about the nature of the roots—whether they are real or complex, and if real, whether they are distinct or repeated.