write an equation of the line tangent to the graph of ( f(x) = 2x^{3}-x^{2}+3x ) at the point where ( x = 1 ).

Explanation:

Step1: Find the y - coordinate of the point

Substitute (x = 1) into (f(x)=2x^{3}-x^{2}+3x). (f(1)=2(1)^{3}-(1)^{2}+3(1)=2 - 1+3=4). So the point is ((1,4)).

Step2: Find the derivative of (f(x))

Using the power rule ((x^{n})^\prime=nx^{n - 1}), (f^\prime(x)=(2x^{3}-x^{2}+3x)^\prime=6x^{2}-2x + 3).

Step3: Find the slope of the tangent line

Substitute (x = 1) into (f^\prime(x)). (f^\prime(1)=6(1)^{2}-2(1)+3=6 - 2+3=7). The slope (m = 7).

Step4: Use the point - slope form (y - y_{1}=m(x - x_{1}))

Here (x_{1}=1,y_{1}=4,m = 7). So (y - 4=7(x - 1)). Simplify: (y-4=7x-7), then (y=7x - 3).

Answer:

(y = 7x-3)