A curve, C, has equation y =(2x-3)^5. A point, P, lies on C at (w,-32). Find the value of w and the equation of the tangent of C at point, P in the form y =mx+c.

To find the value of w, let x = w and y = -32. Substitute these values into the equation of the curve, C: y = (2x-3)^5 => -32 = (2(w) - 3 )^5. Note: the symbol, =>, means "implies that." Fifth root both sides: => (-32)^1/5 = 2w -3 => -2 = 2w - 3. Add 3 to both sides: => -2 + 3 = 2w - 3 + 3 => 1 = 2w. Divide both sides by 2: => 1/2 = (2w)/2=> 1/2 = w => w =1/2. Therefore the value of w is 1/2.Next, to find the equation of the tangent, differentiate the equation of the curve, C, with respect to x: dy/dx = (5)(2x-3)^4(2) => dy/dx = 10(2x-3)^4. Substitute value x = w = 1/2: dy/dx = 10(2(1/2)-3)^4 => dy/dx = 10(1-3)^4 => dy/dx = 10(-2)^4 => dy/dx = 10(16) => dy/dx = 160. Let dy/dx = m. Using the equation: (y-ya) = m(x-xa), let ya = -32 and xa = 1/2. So (y - (-32)) = 160(x - (1/2)) => y + 32 = 160x - 80. Subtract both sides by 32: y + 32 - 32 = 160x - 80 - 32 => y = 160x - 112.Therefore, the equation of the tangent of the curve, C, at the point, P, is y = 160x - 112.