Question

how can i find a tangent equation to f(x)= √x at x=2

Answer 1

Find the derivative \(f^\prime\) and then use \(f^{\prime}(2)\) as the slope in point-slope form to find the equation of the tangent line.

Answer 2

but what if it doesnt give me a y?

Answer 3

\(f(x) = y\)

Answer 4

So in this case \(y=\sqrt{2}\).

Answer 5

\[\prime=\left(\begin{matrix}1 \\ 2\sqrt{x}\end{matrix}\right)\] but then what do i do

Answer 6

\(f^{\prime}(x) = \frac{1}{2\sqrt{x}}\)
Point-slope form: \(y - f(x_1) = f^{\prime}(x_1) * (x - x_1)\)

Answer 7

so i replace the x in the slope for 2 and then plug in everything else i know for b??

Answer 8

\(x_1 = 2\) in this case.

Answer 9

\(x\) and \(y\) will be the variables in the final equation.

Answer 10

It may be more familiar to you as \(y - y_1 = m(x - x_1)\), it's the same thing.

Answer 11

well i know that but dont i need a b?
or is the final equation \[y=\left(\begin{matrix}1 \\ 2\sqrt{x}\end{matrix}\right)x+b\]

Answer 12

\(y = mx + b\) is slope-intercept form. \(b\) just refers to any constant. When you evaluate point-slope form you will end up with that constant.

Answer 13

\(y - \sqrt{2} = \frac{1}{2\sqrt{2}}\left(x - 2\right)\)

Answer 14

Solving for \(y\) will give you slope-intercept form.

Answer 15

okay thank