Find the equation of the tangent to the curve y=x^2−x+3 at the point (4,15)


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.

Find the equation of the tangent to the curve y = x^2 − x + 3 at the point (4,15).

Discover how to find the equation of the tangent to the curve y = x^2 − x + 3 at the point (4,15). This problem involves differentiation and the point-slope form of a line, perfect for students in Grades 10-12.

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