a) Graph the function
​f(x)equals
x squared plus 6 x
.
​b) Draw the tangent lines to the graph at the points whose​ x-coordinates are
negative 5
​,
negative 4
​,
and
negative 2
.
​c) Find
f prime left parenthesis x right parenthesis
by determining
ModifyingBelow lim With h right arrow 0
StartFraction f left parenthesis x plus h right parenthesis minus f left parenthesis x right parenthesis Over h EndFraction
.
​d) Find
f prime left parenthesis negative 5 right parenthesis
​,
f prime left parenthesis negative 4 right parenthesis
​,
and
f prime left parenthesis negative 2 right parenthesis
.
These slopes should match those of the lines in part​ (b).

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.

a) Graph the function f(x) = x^2 + 6x. b) Draw the tangent lines to the graph at the points whose x-coordinates are -5, -4, and -2. c) Find f'(x) using the limit definition of the derivative. d) Find f'(-5), f'(-4), and f'(-2), and verify that these slopes match those of the tangent lines in part b.

This problem involves graphing the quadratic function f(x) = x^2 + 6x, finding the tangent lines at specified points, and calculating the derivative using the limit definition. Suitable for Grades 11-12 students learning about derivatives and their applications.

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