Question

How do you find the equation of the line tangent to the graph of $y = {x^2}$ at the point $x = 1$ ?

Answer 1

In this question, we have to find out the equation of the tangent line with the help of the given equation and the given point. First, differentiate the given equation to get the slope. Then, put $x = 1$ in the given equation to get the value of y-coordinate. Now, we have a point and a slope. Put the information in point-slope form. And you will get your answer.

Formula used: To find the slope m of a curve at a particular point, we differentiate the equation of the curve. If the given curve is $y = f\left( x \right)$ we evaluate $\dfrac{{dy}}{{dx}}$ or $f'\left( x \right)$ and substitute the value of x to find the slope.
Then the tangent at $\left( {{x_1},{y_1}} \right)$ is $y - {y_1} = m\left( {x - {x_1}} \right)$ .
Differentiation formula: $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$ .

Complete step-by-step solution:
We have to find out the tangent line to the curve $y = {x^2}$ at the point $x = 1$ .
Here, $x = 1$, thus $y = {\left( 1 \right)^2} = 1$ .
Hence, the point at which slope is to be found is $\left( {1,1} \right)$.
To find the slope m of a curve at a particular point, we need to differentiate the equation of the curve.
Differentiating the given curve, we have,
$\dfrac{{dy}}{{dx}} = 2x$
Hence the slope of the tangent line is
${\left. {\dfrac{{dy}}{{dx}}} \right|_{x = 1}} = 2 \times 1 = 2$ .
Using the formula of the equation of the tangent line at $\left( {{x_1},{y_1}} \right)$ is $y - {y_1} = m\left( {x - {x_1}} \right)$ , we get,
The equation of the tangent line to the curve $y = {x^2}$ at point $\left( {1,1} \right)$ is
$\Rightarrow y - 1 = 2\left( {x - 1} \right)$
Shifting the terms to find a simplified equation,
$\Rightarrow y - 1 = 2x - 2$
$\Rightarrow 2x - y = 2 - 1$
$\Rightarrow 2x - y = 1$

Hence, the equation of the line tangent to the graph of $y = {x^2}$ at the point $x = 1$ is $2x - y = 1$ .

Note: To determine the equation of a tangent to a curve:
Find the derivative using the rules of differentiation.
Substitute the x-coordinate and the y-coordinate of the given point into the derivative to calculate the slope of the tangent.
Substitute the slope of the tangent and the coordinates of the given point into an appropriate form of the straight- line equation, i.e., $y - {y_1} = m\left( {x - {x_1}} \right)$ where m is the slope of the tangent and $\left( {{x_1},{y_1}} \right)$ is the given point.