To find the equation of the tangent line to the curve $y = x^2 + 3$ at the point $(2,7)$, follow these steps:

Step 1: Differentiate the Function

The derivative of $y = x^2 + 3$ represents the slope of the tangent line:

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

Step 2: Evaluate the Slope at $x = 2$

$m = 2(2) = 4$

Thus, the slope of the tangent line at $(2,7)$ is $m = 4$.

Step 3: Use the Point-Slope Form

The equation of a tangent line is given by:

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

Substituting $m = 4$ and $(x_1, y_1) = (2,7)$:

$y - 7 = 4(x - 2)$

Step 4: Simplify

$y - 7 = 4x - 8$

$y = 4x - 1$

Final Answer:

$\boxed{y = 4x - 1}$

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Related Questions:

1. Find the equation of the tangent line to $y = x^3 - 2x + 5$ at $x = 1$.
2. Determine the normal line equation to $y = x^2 + 3$ at $x = 2$.
3. Find the derivative of $y = \sqrt{x} + 2x$ and determine the slope at $x = 4$.
4. Compute the second derivative of $y = x^2 + 3$.
5. Find the points where the tangent to $y = x^2 - 4x + 4$ is horizontal.

Tip:

The normal line at a point on a curve is perpendicular to the tangent line and has a slope that is the negative reciprocal of the tangent line's slope.