Question

How do you find the equation of the tangent line to the curve $3x^2 -5x +2$ at x = 3 ?

Answer 1

To do this question, first, you need to find the slope of the curve. To do that you need to differentiate the curve and find the value of the y’ at the point x = 3. Then, you need to find the value of y for the original equation at x = 3. Then you get the point (3 , 14) as x = 3. Now use the equation $y – y_1 = m\left( x – x_1 \right)$. Here, substitute m as 13. $x_1$ as 3 and $y_1$ as 14. Then you get the tangent line equation.

Complete step by step solution:
Here is the complete step by step solution.
The first step we need to do is to find the slope of the curve. We can find the slope of the curve by differentiating the curve and plugging in the value x = 3 to get the final value for slope. Therefore, we get
$\Rightarrow y = 3x^2 -5x +2$
$\Rightarrow y’ = 6x-5$
$\Rightarrow y’\left( 3 \right) = 13$
Now, we find the value of y for the original equation at x = 3. Then, we get
$\Rightarrow y = 3.3^2 -5.3 +2$
$\Rightarrow y = 14$
Therefore, now get the point (3, 14) and slope 13. Therefore, we find the equation of the tangent line using the equation $y – y_1 = m\left( x – x_1 \right)$. By doing this, we get the equation as:
$y – y_1 = m\left( x – x_1 \right)$
$\Rightarrow y – 14 = 13\left( x – 3 \right)$
$\Rightarrow y – 14 = 13x - 39$
$\Rightarrow y – 13x + 25 = 0$

Therefore, we get the final answer of the question, as $y – 13x + 25 = 0$.

Note: You need to know the different types of lines to find the tangent line like the two points form, or the one point and the slope form which was the one we used above and also the slope and the intercept form.