Let's break down and solve each part of the problem step by step.

Part a) Graph the function $f(x) = x^2 + 6x$.

The function $f(x) = x^2 + 6x$ is a quadratic equation. To graph this function, we can:

1. Identify the key points:

   • Vertex: The vertex of a parabola given by $f(x) = ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$. Here, $a = 1$ and $b = 6$, so: $x_{\text{vertex}} = -\frac{6}{2(1)} = -3$ Substituting $x = -3$ into the function to find the y-coordinate of the vertex: $f(-3) = (-3)^2 + 6(-3) = 9 - 18 = -9$ So, the vertex is at $(-3, -9)$.

2. Intercepts:

   • y-intercept: When $x = 0$: $f(0) = 0^2 + 6(0) = 0$ So, the y-intercept is $(0, 0)$.
   • x-intercepts: Set $f(x) = 0$: $x^2 + 6x = 0 \implies x(x + 6) = 0$ So, $x = 0$ and $x = -6$. The x-intercepts are $(0, 0)$ and $(-6, 0)$.

3. The parabola opens upwards because the coefficient of $x^2$ is positive.

Part b) Draw the tangent lines to the graph at the points whose x-coordinates are $x = -5$, $x = -4$, and $x = -2$.

To draw the tangent lines, we first need the slopes of the tangent lines at these points, which will be calculated in part d).

Part c) Find $f'(x)$ by determining:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ Start by calculating $f(x+h)$: $f(x+h) = (x+h)^2 + 6(x+h) = x^2 + 2xh + h^2 + 6x + 6h$ Now, find $f(x+h) - f(x)$: $f(x+h) - f(x) = (x^2 + 2xh + h^2 + 6x + 6h) - (x^2 + 6x) = 2xh + h^2 + 6h$ Now, divide by $h$: $\frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2 + 6h}{h} = 2x + h + 6$ Taking the limit as $h \to 0$, we get: $f'(x) = 2x + 6$ So, the derivative of the function is: $f'(x) = 2x + 6$

Part d) Find $f'(-5)$, $f'(-4)$, and $f'(-2)$.

Using the derivative $f'(x) = 2x + 6$, we can find the slopes of the tangent lines at the specified points:

1. $f'(-5) = 2(-5) + 6 = -10 + 6 = -4$
2. $f'(-4) = 2(-4) + 6 = -8 + 6 = -2$
3. $f'(-2) = 2(-2) + 6 = -4 + 6 = 2$

These slopes correspond to the slopes of the tangent lines at $x = -5$, $x = -4$, and $x = -2$.

Summary of results:

• Part a: The graph is a parabola with vertex at $(-3, -9)$, and it opens upward.
• Part b: The tangent lines have slopes of $-4$, $-2$, and $2$ at the points $x = -5$, $x = -4$, and $x = -2$ respectively.
• Part c: The derivative of the function is $f'(x) = 2x + 6$.
• Part d: The slopes of the tangent lines are $f'(-5) = -4$, $f'(-4) = -2$, and $f'(-2) = 2$.

Would you like a more detailed graph, or do you have any questions? Here are five related questions to expand your understanding:

1. How do you interpret the meaning of the slope of a tangent line in real-world applications?
2. What are the implications of the sign of the derivative for the shape of a graph?
3. How would the graph change if the coefficient of $x^2$ were negative?
4. What happens to the derivative $f'(x)$ as $x$ approaches positive or negative infinity?
5. How can you estimate the derivative of a function graphically?

Tip: When finding a derivative using the limit definition, remember to simplify the expression before taking the limit to avoid indeterminate forms.