What is the equation of the line tangent to f(x)=2x3−x2+3x at x=3?

Question

Branden Nelson · Accepted Answer

Explanation:We calculate the first derivativef(x)=2x3−x2+3xf′(x)=6x2−2x+3The equation of the tangent isy-f(3)=f&#039;(3)(x-3)f(3)=2⋅33−32+3⋅3=54−9+9=54f′(3)=6⋅32−2⋅3+3=54−6+3=51The tangent isy-54=51(x-3)y=51x-99

Kaylie Carroll · Accepted Answer

When finding the tangent/normal to a line, you always have to find the gradient first.
The equation:

f (x) = 2 x^{3} - x^{2} + 3 x

when differentiated will become

f^{'} (x) = 6 x^{2} - 2 x + 3 .

At x=3, the gradient of the curve and the tangent will be 51.
To find the equation of the tangent, you need to use the format

y - y_{1} = m (x - x_{1})

with m being your gradient and

x_{1} y_{1}

being your x and y values. We have the x value already, so we need to find the y value by substituting the x value into the original equation.

2 \cdot 3^{3} - 3^{2} + 3 \cdot 3 = 54

At (3 , 54) the equation of the tangent would be:
y-54=51(x-3)
Simplify it by expanding the brackets on the right and the final answer will be
y=51x-99

What is the equation of the line tangent to f(x)=2x^{3}-x^{2}+3x

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