24. [-/1 Points] DETAILS MY NOTES SCALCET9 3.XP.1.038. Find an equation of the tangent line to the curve \( y=x \sqrt{x} \) that is parallel to the line \( y=6+9 x \). \( y=\square \)

To find the equation of the tangent line to the curve \( y = x \sqrt{x} \) that is parallel to the line \( y = 6 + 9x \), we first note that the slope of the given line is 9. We need to find a point on the curve where the derivative (slope) is also 9. The curve can be rewritten as \( y = x^{3/2} \). We differentiate this equation: \[ \frac{dy}{dx} = \frac{3}{2} x^{1/2}. \] Setting this equal to 9 gives us: \[ \frac{3}{2} x^{1/2} = 9 \] which leads to: \[ x^{1/2} = 6 \] and squaring both sides results in \( x = 36 \). Now we substitute \( x = 36 \) back into the equation of the curve to find the corresponding \( y \)-value: \[ y = 36 \sqrt{36} = 36 \times 6 = 216. \] Now we have the point \( (36, 216) \). The tangent line has the form: \[ y - 216 = 9(x - 36). \] Rearranging gives the equation of the tangent line: \[ y = 9x - 324 + 216 \] or: \[ y = 9x - 108. \]