\(y = \frac{1}{2} x^4 -x^3 +5x \)

\(\frac{dy}{dx} = 2x^3 -3x^2 +5\)

The slopes of the tangent lines is \(m = \tan 45 = 1\) and \(m = \tan 135 = -1\)

Find the points of tangency along the curve.

\(\frac{dy}{dx} = 2x^3 -3x^2 +5 = 1\)

\( x_1 \approx -0.9108200822\)

\( x_2 \approx -1.0786168851 \)

\(y\) values

\(y_1 = \frac{1}{2} (-0.91)^4 -(-0.91)^3 +5(-0.91)\)

\(y_1 = -3.45437778711\)

\(y_2 = \frac{1}{2} (-1.08)^4 -(-1.08)^3 +5(-1.08)\)

\(y_2 = -3.46143951026\)

Thus the points of tangency are \((-0.91, -3.45) ~ and ~ (-1.08, -3.46)\)

Find the first tangent.

\(y--3.45 = 1 (x--0.91) \)

\(y--3.45 = x+0.91 \)

\(y = x -2.54\)

The second tangent.

\(y--3.46 =-1(x--1.08) \)

\(y--3.46 = -x-1.08\)

\(y = -x -4.54\)