To find the equation of the tangent to the curve $y = x^2 - x + 3$ at the point $(4, 15)$, we need to follow these steps:

Step 1: Differentiate the equation to find the slope of the tangent line.

The slope of the tangent line is given by the derivative of $y$ with respect to $x$.

$\frac{dy}{dx} = \frac{d}{dx}(x^2 - x + 3)$ $\frac{dy}{dx} = 2x - 1$

Step 2: Find the slope at the point $(4, 15)$.

Substitute $x = 4$ into the derivative to get the slope at that point:

$\text{slope} = 2(4) - 1 = 8 - 1 = 7$

Step 3: Use the point-slope form of the equation of a line.

The point-slope form is:

$y - y_1 = m(x - x_1)$

Where:

• $(x_1, y_1) = (4, 15)$ is the point of tangency,
• $m = 7$ is the slope.

Substitute these values:

$y - 15 = 7(x - 4)$

Step 4: Simplify the equation.

Expand and simplify:

$y - 15 = 7x - 28$ $y = 7x - 28 + 15$ $y = 7x - 13$

Final Answer:

The equation of the tangent to the curve at the point $(4, 15)$ is:

$y = 7x - 13$

Would you like further details on any step or have any questions?

Here are some related questions for practice:

1. What is the derivative of $y = 3x^2 - 2x + 5$?
2. How would you find the equation of a normal line to a curve?
3. Can the tangent line at a point on a curve be horizontal? If so, when?
4. Find the slope of the tangent line to the curve $y = x^3 - x$ at $x = 2$.
5. How do you interpret the second derivative of a curve geometrically?

Tip: The slope of the tangent line gives the instantaneous rate of change of the curve at that specific point.