Tangent Line Properties at a Point

Finding the Slope of the Tangent

To determine the properties of the tangent line to the curve defined by the equation $y = 5x^2 + 3$ at the point $(0, 3)$, we first need to find the slope of the tangent at this specific point.

The slope of the tangent line at any point on a curve is given by the derivative of the function, $\frac{dy}{dx}$.

First, find the derivative of $y = 5x^2 + 3$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(5x^2 + 3)$

$\frac{dy}{dx} = 10x$

Evaluating the Slope at x = 0

Now, substitute the x-coordinate of the given point, $x = 0$, into the derivative to find the slope ($m$) at that point:

$m = \frac{dy}{dx}\Big|_{x=0} = 10(0)$

$m = 0$

Interpreting the Slope

A slope of $0$ indicates that the tangent line is horizontal.

A horizontal line is parallel to the x-axis.

Analyzing the Options

Based on the calculated slope:

• Option 1: The tangent equation is $y - 3 = 0(x - 0)$, which is $y = 3$. This line does not pass through $(0, 0)$.
• Option 2: The slope is 0, not +1.
• Option 3: The slope is 0, meaning the tangent line is parallel to the x-axis. This is consistent with our findings.
• Option 4: The slope is 0, not -1.

Therefore, the tangent line at the point $(0, 3)$ is parallel to the x-axis.