Let's work through this problem step-by-step.

We are given the function:

$f(x) = 2x^2 - 4x + 9$

We are tasked with finding:

1. The slope of the tangent line to the graph of $f(x)$ at the point $(3, 15)$.
2. The equation of the tangent line in the form $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

Step 1: Find the slope of the tangent line

The slope of the tangent line is the derivative of $f(x)$ evaluated at $x = 3$. To find the derivative $f'(x)$, we differentiate $f(x) = 2x^2 - 4x + 9$:

$f'(x) = \frac{d}{dx}(2x^2) - \frac{d}{dx}(4x) + \frac{d}{dx}(9)$

$f'(x) = 4x - 4$

Now, evaluate $f'(x)$ at $x = 3$:

$f'(3) = 4(3) - 4 = 12 - 4 = 8$

Thus, the slope of the tangent line at $(3, 15)$ is 8.

Step 2: Find the equation of the tangent line

The equation of a line is given by:

$y = mx + b$

where $m$ is the slope, and $b$ is the y-intercept. We already know that $m = 8$, and we can use the point $(3, 15)$ to find $b$.

Substitute $x = 3$, $y = 15$, and $m = 8$ into the equation:

$15 = 8(3) + b$

$15 = 24 + b$

$b = 15 - 24 = -9$

Thus, the equation of the tangent line is:

$y = 8x - 9$

Final Answers:

• The slope of the tangent line is 8.
• The equation of the tangent line is $y = 8x - 9$.

Would you like further clarification or more details? Here are five related questions to deepen your understanding:

1. What is the geometric interpretation of the tangent line in this context?
2. How would the equation change if the point of tangency was at $x = 2$ instead?
3. Can you verify the value of $f(3)$ to ensure the point $(3, 15)$ lies on the graph of $f(x)$?
4. How does the second derivative $f''(x)$ help describe the curvature of the graph near $x = 3$?
5. How would you use the same process to find the tangent line for a cubic function like $f(x) = x^3 - 2x + 1$?

Tip: The derivative gives the slope of the tangent line at any point on the curve, which is key in analyzing how the function behaves locally.