# Differentiate and find the equation of the tangent and the normal. `y=sqrt(2x+3) ` at the point x=3

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### 1 Answer

First, plug-in x=3 to the equation of the curve to get the corresponding value of y.

`y=sqrt(2x+3)=sqrt(2*3+3)=sqrt9=3`

So, the point of tangency is (3, 3).

Next, determine the slopes of the tangent and the normal line.

To do so, take the derivative of the given curve.

`y=sqrt(2x+3)`

`y'=(sqrt(2x+3))'`

To get the derivative, express the radical as exponent and apply the power rule which is `(u^n)'=n*u^(n-1)*u'` .

`y'=((2x+3)^(1/2))'`

`y'=1/2(2x+3)^(-1/2) *(2x+3)'`

`y'=1/2(2x+3)^(-1/2)*2`

`y'=(2x+3)^(-1/2)`

To simplify, apply the negative exponent rule which is `a^(-m)=1/a^m` .

`y'=1/(2x+3)^(1/2)`

`y'=1/(2x+3)^(1/2)`

`y'=1/(sqrt(2x+3))`

Note that the slope of a tangent at a certain point in the curve is equal to y' `( m_T=y')` .

So, plug-in x=3 to y'.

`y'=1/(sqrt(2*3+3))=1/(sqrt9)=1/3=1/3`

Hence, the slope of the tangent line at point (3,3) is `m_T=1/3` .

Since a normal line is perpendicular to tangent line, its slope is:

`m_N=-1/m_T=-1/(1/3)=-3`

Now that the point of tangency and the slopes of the two lines are known, use the point-slope formula to get the equation of each line.

The formula is:

`y-y_1=m(x-x_1)`

For the tangent line, plug-in `m_T=1/3` and the point (3,3) to the formula.

`y-3=1/3(x-3)`

`y-3=x/3-3/3`

`y-3=x/3-1`

`y-3+3=x/3-1+3`

`y=x/3+2`

**Hence, the equation of the tangent line is `y=x/3+2` .**

For the normal line, substitute `m_N=-3` and the point (3,3) to the same formula.

`y-3=-3(x-3)`

`y-3=-3x+9`

`y-3+3=-3x+9+3`

`y=-3x+12`

**Thus, the equation of the normal line is `y=-3x+12` .**