Question

Find an equation of the tangent line to the curve at the given point.

37. \(y = 2{x^3} - {x^2} + 2,\;\;\;\;\;\;\;\left( {1,3} \right)\)

Short answer

The required equation is \(y = 4x - 1\).

Step-by-step solution

Find the slope of the equation

Determine the derivative \(\frac{{dy}}{{dx}}\) to represent the slope.

\(\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {2{x^3} - {x^2} + 2} \right)\\ &= \frac{d}{{dx}}\left( {2{x^3}} \right) - \frac{d}{{dx}}\left( {{x^2}} \right) + \frac{d}{{dx}}\left( 2 \right)\\ &= 2\left( {3{x^{3 - 1}}} \right) - 2{x^{2 - 1}} + 0\\y' &= 6{x^2} - 2x\end{align}\)

Find the slope of the equation at given point

Substitute the given point into the obtained equation of slope to find the slope at particular point.

\(\begin{align}y'\left( 1 \right) &= 6{\left( 1 \right)^2} - 2\left( 1 \right)\\ &= 6\left( 1 \right) - 2\\ &= 6 - 2\\ &= 4\end{align}\)

Determine the tangent line equation

The equation of tangent line is given by\(y - {y_1} = m\left( {x - {x_1}} \right)\), where\(\left( {{x_1},{y_1}} \right)\)is the given pint and\(m\)is the slope at given point.

Substitute the values into the equation to find the equation of tangent line at the point \(\left( {1,3} \right)\).

\(\begin{align}y - 3 &= 4\left( {x - 1} \right)\\y - 3 &= 4x - 4\\y &= 4x - 4 + 3\\y &= 4x - 1\end{align}\)

Thus, the required equation is \(y = 4x - 1\).