Find an equation of the tangent line to the curve at the given point. \[ y=7 x-4 \sqrt{x},(1,3) \] \( y=\square \)

**Step 1. Differentiate the function** The function is given by \[ y = 7x - 4 \sqrt{x}. \] We rewrite \( \sqrt{x} \) as \( x^{\frac{1}{2}} \) to differentiate: \[ y = 7x - 4x^{\frac{1}{2}}. \] Differentiate term-by-term: - The derivative of \(7x\) is \(7\). - For \(-4x^{\frac{1}{2}}\), use the power rule: \[ \frac{d}{dx} [x^{\frac{1}{2}}] = \frac{1}{2}x^{-\frac{1}{2}}, \] so the derivative is \[ -4 \cdot \frac{1}{2} x^{-\frac{1}{2}} = -2x^{-\frac{1}{2}} = -\frac{2}{\sqrt{x}}. \] Thus, the derivative of the function is \[ y' = 7 - \frac{2}{\sqrt{x}}. \] **Step 2. Evaluate the derivative at \( x = 1 \)** Substitute \( x = 1 \) into the derivative: \[ y'(1) = 7 - \frac{2}{\sqrt{1}} = 7 - 2 = 5. \] This value represents the slope \( m \) of the tangent line at the point \((1,3)\). **Step 3. Write the equation of the tangent line** Using point-slope form: \[ y - y_1 = m(x - x_1), \] with \( (x_1, y_1) = (1,3) \) and \( m = 5 \), we have: \[ y - 3 = 5(x - 1). \] Solve for \( y \): \[ y = 5x - 5 + 3 = 5x - 2. \] Thus, the equation of the tangent line is \[ y = 5x - 2. \]