Answer:
[tex]y = 4[/tex]
Step-by-step explanation:
First, find the derivative of f:
[tex]\displaystyle \begin{aligned} f'(x) & = \frac{d}{dx}\left[ 2x^3 - 3x^2 + 5\right] \\ \\ & = 6x^2 - 6x \\ \\ & = 6x(x-1)\end{aligned}[/tex]
Find the instantaneous slope at x = 1:
[tex]\displaystyle \begin{aligned} f'(1) & = 6(1)((1)-1) \\ \\ &= 0 \end{aligned}[/tex]
From the point-slope form, find the equation of the line:
[tex]\displaystyle \begin{aligned}y - y_1 & = m(x-x_1) \\ \\ y - (4) & =(0)(x-(1)) \\ \\ y & = 4 \end{aligned}[/tex]
In conclusion, the equation of the tangent line is y = 4.
