Question

find the tangent line to ( y = sqrt{x^{2}-x + 19} ) at ( x = 3 ).
the tangent line to ( y = sqrt{x^{2}-x + 19} ) at ( x = 3 ) is
(type an equation.)

Explanation:

Step1: Find the value of \(y\) when \(x = 3\)

Substitute \(x = 3\) into \(y=\sqrt{x^{2}-x + 19}\).
\(y=\sqrt{3^{2}-3 + 19}=\sqrt{9-3 + 19}=\sqrt{25}=5\)

Step2: Differentiate \(y\) with respect to \(x\)

Let \(u=x^{2}-x + 19\), then \(y = \sqrt{u}=u^{\frac{1}{2}}\).
Using the chain - rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\).
\(\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}\) and \(\frac{du}{dx}=2x - 1\).
So \(\frac{dy}{dx}=\frac{2x - 1}{2\sqrt{x^{2}-x + 19}}\)

Step3: Find the slope of the tangent line at \(x = 3\)

Substitute \(x = 3\) into \(\frac{dy}{dx}\).
\(\frac{dy}{dx}\big|_{x = 3}=\frac{2\times3-1}{2\sqrt{3^{2}-3 + 19}}=\frac{6 - 1}{2\times5}=\frac{5}{10}=\frac{1}{2}\)

Step4: Use the point - slope form \(y - y_{1}=m(x - x_{1})\)

Here \(x_{1}=3,y_{1}=5,m=\frac{1}{2}\).
\(y - 5=\frac{1}{2}(x - 3)\)
Expand: \(y-5=\frac{1}{2}x-\frac{3}{2}\)
\(y=\frac{1}{2}x-\frac{3}{2}+5\)
\(y=\frac{1}{2}x+\frac{7}{2}\)

Answer:

\(y=\frac{1}{2}x+\frac{7}{2}\)