To draw the graph of the quadratic function $f(x) = 6x^2 + 8x + 2$, we can follow these steps:

1. Identify the coefficients:

   • The function is in the form $f(x) = ax^2 + bx + c$, where:
     • $a = 6$
     • $b = 8$
     • $c = 2$

2. Find the vertex using the formula for the x-coordinate of the vertex of a parabola, which is: $x = \frac{-b}{2a}$ Plugging in the values of $a$ and $b$: $x = \frac{-8}{2(6)} = \frac{-8}{12} = -\frac{2}{3}$ To find the y-coordinate of the vertex, substitute $x = -\frac{2}{3}$ into the function: $f\left(-\frac{2}{3}\right) = 6\left(-\frac{2}{3}\right)^2 + 8\left(-\frac{2}{3}\right) + 2$ Simplifying the calculation: $f\left(-\frac{2}{3}\right) = 6\left(\frac{4}{9}\right) - \frac{16}{3} + 2$ $f\left(-\frac{2}{3}\right) = \frac{24}{9} - \frac{16}{3} + 2 = \frac{8}{3} - \frac{16}{3} + 2 = -\frac{8}{3} + 2 = -\frac{8}{3} + \frac{6}{3} = -\frac{2}{3}$ So, the vertex is at $\left(-\frac{2}{3}, -\frac{2}{3}\right)$.

3. Find the axis of symmetry: The axis of symmetry is the vertical line through the vertex. Its equation is: $x = -\frac{2}{3}$

4. Find the y-intercept by setting $x = 0$ in the function: $f(0) = 6(0)^2 + 8(0) + 2 = 2$ So, the y-intercept is $(0, 2)$.

5. Plot a few more points by selecting additional values for $x$ and calculating the corresponding values of $f(x)$.

I will now generate a graph for this function. numpy as np import matplotlib.pyplot as plt

Define the function

def f(x): return 6x**2 + 8x + 2

Generate x values

x = np.linspace(-5, 3, 400) y = f(x)

Plot the function

plt.figure(figsize=(8, 6)) plt.plot(x, y, label=r'$f(x) = 6x^2 + 8x + 2$', color='b') plt.axhline(0, color='black',linewidth=1) plt.axvline(0, color='black',linewidth=1)

Mark the vertex (-2/3, -2/3)

vertex_x = -2/3 vertex_y = f(vertex_x) plt.plot(vertex_x, vertex_y, 'ro') # Red dot for vertex plt.text(vertex_x, vertex_y, f' ({vertex_x:.2f}, {vertex_y:.2f})', verticalalignment='bottom')

Mark the y-intercept (0, 2)

plt.plot(0, 2, 'go') # Green dot for y-intercept plt.text(0, 2, ' (0, 2)', verticalalignment='bottom')

Set the axis limits and labels

plt.xlim(-5, 3) plt.ylim(-10, 40) plt.title("Graph of the function $f(x) = 6x^2 + 8x + 2$") plt.xlabel("x") plt.ylabel("f(x)") plt.grid(True) plt.legend()

Show plot

plt.show()Here is the graph of the quadratic function $f(x) = 6x^2 + 8x + 2$.

• The vertex is located at $\left(-\frac{2}{3}, -\frac{2}{3}\right)$.
• The y-intercept is at $(0, 2)$.
• The parabola opens upwards because the coefficient of $x^2$ (which is 6) is positive.

Would you like more details about the graph or any other related questions?

Here are some related questions for further exploration:

1. How do we find the x-intercepts (roots) of a quadratic function?
2. What is the role of the discriminant in determining the number of real roots of a quadratic equation?
3. How does the coefficient of $x^2$ affect the width of the parabola?
4. Can you explain how to derive the vertex form of this quadratic function?
5. How do we sketch the graph of a quadratic equation by hand?

Tip: A good way to sketch quadratic functions by hand is to first identify the vertex, axis of symmetry, and a few points by plugging in values for $x$, and then reflect those points across the axis of symmetry.