Question

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

5. \(y = 2{x^2} - 5x + 1\) , \(\left( {3,4} \right)\)

Short answer

The equation of the tangent line to the curve at the given point is \(y = 7x - 17\).

Step-by-step solution

The slope of the tangent to the curve

The formula of the slope of the tangent line to the curve is defined as:

\(m = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}}\)

The slope of the tangent to the given curve

Substitute\(a = 3\)in the above formula to obtain the slope as follows:

\(\begin{aligned}m &= \mathop {\lim }\limits_{x \to 3} \frac{{f\left( x \right) - f\left( 3 \right)}}{{x - 3}}\\ &= \mathop {\lim }\limits_{x \to 3} \frac{{2{x^2} - 5x + 1 - \left( {2{{\left( 3 \right)}^2} + 5\left( 3 \right) + 1} \right)}}{{x - 3}}\\ &= \mathop {\lim }\limits_{x \to 3} \frac{{2{x^2} - 5x - 3}}{{x - 3}}\end{aligned}\)

Solve further,

\(\begin{aligned}m &= \mathop {\lim }\limits_{x \to 3} \frac{{\left( {2x + 1} \right)\left( {x - 3} \right)}}{{x - 3}}\\ &= \mathop {\lim }\limits_{x \to 3} \left( {2x + 1} \right)\\ &= 2\left( 3 \right) + 1\\ &= 7\end{aligned}\)

Thus, the slope is \(m = 7\).

The equation of a line

The general equation of a line that passes through the point\(\left( {{x_1},{y_1}} \right)\)and has slope\(m\)is given below:

\(y - {y_1} = m\left( {x - {x_1}} \right)\)

The equation of the tangent line

Substitute the value of\({x_1}\),\({y_1}\)in the above formula to obtain the equation of tangent as follows:

\(\begin{aligned}y - 4 &= 7\left( {x - 3} \right)\\y - 4 &= 7x - 21\\y &= 7x - 17\end{aligned}\)

Thus, the equation of the tangent line is \(y = 7x - 17\).