Questions: The slope of the tangent line to the curve y=3/x at the point (4, 3/4) is: -3/16 The equation of this tangent line can be written in the form y=mx+b where: m is: -3/16 b is: 3/2

Solution Steps

To find the equation of the tangent line to the curve \( y = \frac{3}{x} \) at the point \( \left(4, \frac{3}{4}\right) \), we use the point-slope form of the equation of a line. Given the slope \( m = -\frac{3}{16} \) and the point \( (4, \frac{3}{4}) \), we can substitute these values into the point-slope form equation \( y - y_1 = m(x - x_1) \) and solve for \( b \).

We are given the curve \( y = \frac{3}{x} \) and the point \( \left(4, \frac{3}{4}\right) \). The slope of the tangent line at this point is \( m = -\frac{3}{16} \).

The point-slope form of the equation of a line is: \[ y - y_1 = m(x - x_1) \] Substituting the given point \( (4, \frac{3}{4}) \) and the slope \( m = -\frac{3}{16} \): \[ y - \frac{3}{4} = -\frac{3}{16}(x - 4) \]

Rearrange the equation to the slope-intercept form \( y = mx + b \): \[ y = -\frac{3}{16}x + b \] To find \( b \), substitute \( x = 4 \) and \( y = \frac{3}{4} \): \[ \frac{3}{4} = -\frac{3}{16} \cdot 4 + b \] \[ \frac{3}{4} = -\frac{3}{4} + b \] \[ b = \frac{3}{4} + \frac{3}{4} \] \[ b = \frac{6}{4} \] \[ b = 1.5 \]

Final Answer