Questions: Find an equation of the line that is tangent to the graph of f for the given value of x. f(x) = sqrt(3x+16); x=0

Transcript text: Find an equation of the line that is tangent to the graph of $f$ for the given value of $x$. \[ f(x)=\sqrt{3 x+16} ; x=0 \]

Solution

Step 1: Calculate the derivative of $f(x)$

The derivative of $f(x) = \sqrt{3 x + 16}$ is $f'(x) = \frac{3}{2 \sqrt{3 x + 16}}$.

Step 2: Evaluate $f'(x_0)$

The slope of the tangent line at $x = 0$ is $m = 0.375$.

Step 3: Calculate $f(x_0)$

The y-coordinate of the point on the graph of $f(x)$ at $x = 0$ is $f(x_0) = 4$.

Step 4: Use the point-slope form of a line

The equation of the tangent line in point-slope form is $y - 4 = \frac{3 x}{8}$.

Step 5: Simplify the equation

The simplified equation of the tangent line is $3 x - 8 y = -32$.

The equation of the tangent line to the graph of $f(x) = \sqrt{3 x + 16}$ at $x = 0$, rounded to 2 decimal places, is $3 x - 8 y = -32$.

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