Answer:
[tex]y=\frac{ln(3)}{3}x + \frac{1-ln(3)}{3}[/tex]
Step-by-step explanation:
We must find the slope, (m), of the tangent line by checking the value of the derivate at x=1. Then we must solve for the intercept (b)
We will use this derivative rule:
[tex]\frac{d}{dx}a^{u} = a^{u}\cdot ln(a) \cdot \frac{du}{dx}[/tex]
Take the derivative:
[tex]\frac{d}{dx}y = \frac{d}{dx}3^{x-2}[/tex][tex]\frac{dy}{dx}=3^{x-2}\cdot ln(3) \cdot 1[/tex]
[tex]\frac{dy}{dx} = 3^{x-2} \cdot ln(3) \cdot \frac{d}{dx}(x-2)[/tex]
[tex]\frac{dy}{dx}=3^{x-2}\cdot ln(3)[/tex]
Plug in x=1 to the derivative:
[tex]3^{1-2}\cdot ln(3) = \frac{ln(3)}{3}[/tex]
This is the slope of the tangent.
The original equation at x=1 is: [tex]3^{1-2}=\frac{1}{3}[/tex]
Solve for b:
[tex]\frac{1}{3} =\frac{ln(3)}{3} + b\\b=\frac{1-ln(3)}{3}[/tex]
Now put the slope and the intercept together into a final equation:
[tex]y=\frac{ln(3)}{3}x + \frac{1-ln(3)}{3}[/tex]
Answer:
[tex] \boxed{ \sf y = \left(\frac{1}{3} \cdot \ln(3)\right)x + \left(\frac{1}{3} - \frac{1}{3} \cdot \ln(3)\right)} [/tex]
Step-by-step explanation:
To find the equation of the tangent line to the graph of the function [tex] y = 3^{(x-2)} [/tex] at [tex] x = 1 [/tex], we need to follow these steps:
1. Find the derivative of the function with respect to [tex] x [/tex] to get the slope of the tangent line.
2. Evaluate the derivative at [tex] x = 1 [/tex] to find the slope at that point.
3. Use the point where [tex] x = 1 [/tex] to find the [tex] y [/tex]-coordinate on the original function.
4. Use the point-slope form of the line to find the equation of the tangent line.
Step 1: Derivative of the Function
The derivative of [tex] y = 3^{(x-2)} [/tex] with respect to [tex] x [/tex] can be found using the chain rule, where the derivative of [tex] 3^u [/tex] with respect to [tex] u [/tex] is [tex] 3^u \ln(3) [/tex], and [tex] u = x - 2 [/tex].
So the derivative (denoted as [tex] y' [/tex]) is:
[tex] y' = \dfrac{d}{dx} (3^{(x-2)}) = 3^{(x-2)} \cdot \ln(3) \cdot \dfrac{d}{dx} (x-2) [/tex]
[tex] y' = 3^{(x-2)} \cdot \ln(3) \cdot (1) [/tex]
[tex] y' = 3^{(x-2)} \cdot \ln(3) [/tex]
Step 2: Evaluate the Derivative at [tex] x = 1 [/tex]
[tex] y'(1) = 3^{(1-2)} \cdot \ln(3) [/tex]
[tex] y'(1) = 3^{-1} \cdot \ln(3) [/tex]
[tex] y'(1) = \frac{1}{3} \cdot \ln(3) [/tex]
Step 3: Find the [tex] y [/tex]-Coordinate at [tex] x = 1 [/tex]
[tex] y(1) = 3^{(1-2)} [/tex]
[tex] y(1) = 3^{-1} [/tex]
[tex] y(1) = \frac{1}{3} [/tex]
Step 4: Equation of the Tangent Line in Point-Slope Form
We'll use the point [tex] (1, \frac{1}{3}) [/tex] and the slope [tex] m = \frac{1}{3} \cdot \ln(3) [/tex] to write the equation of the tangent line in point-slope form:
[tex] y - y_1 = m(x - x_1) [/tex]
Plugging in the values:
[tex] y - \frac{1}{3} = \left(\frac{1}{3} \cdot \ln(3)\right)(x - 1) [/tex]
To express it in slope-intercept form ( [tex] y = mx + b [/tex] ), we'll expand the equation and solve for [tex] y [/tex]:
[tex] y = \left(\frac{1}{3} \cdot \ln(3)\right)x - \left(\frac{1}{3} \cdot \ln(3)\right)(1) + \frac{1}{3} [/tex]
[tex] y = \left(\frac{1}{3} \cdot \ln(3)\right)x - \frac{1}{3} \cdot \ln(3) + \frac{1}{3} [/tex]
To simplify, notice that the [tex] -\frac{1}{3} \cdot \ln(3) + \frac{1}{3} [/tex] is a single term, and we can combine like terms:
[tex] y = \left(\frac{1}{3} \cdot \ln(3)\right)x + \left(\frac{1}{3} - \frac{1}{3} \cdot \ln(3)\right) [/tex]
The term inside the parenthesis is the [tex] y [/tex]-intercept ([tex] b [/tex]).
Hence, the equation of the tangent line to the graph of [tex] y = 3^{(x-2)} [/tex] at [tex] x = 1 [/tex] in slope-intercept form is:
[tex] y = \left(\frac{1}{3} \cdot \ln(3)\right)x + \left(\frac{1}{3} - \frac{1}{3} \cdot \ln(3)\right) [/tex]
Learn more about slope-intercept form here: brainly.com/question/35415252
