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To find the equation of the tangent line to the graph of [tex]\( f(x) = \sqrt{x - 9} \)[/tex] at the point [tex]\((13, 2)\)[/tex], we need to follow several steps: find the derivative of [tex]\( f(x) \)[/tex], evaluate the derivative at [tex]\( x = 13 \)[/tex] to determine the slope, and then use the point-slope form of the equation of a line.
### Step 1: Find the derivative of [tex]\( f(x) \)[/tex]
The function given is:
[tex]\[ f(x) = \sqrt{x - 9} \][/tex]
First, express the function as:
[tex]\[ f(x) = (x - 9)^{1/2} \][/tex]
To find the derivative of [tex]\( f(x) \)[/tex], use the power rule and the chain rule:
[tex]\[ f'(x) = \frac{1}{2}(x - 9)^{-1/2} \cdot (1) \][/tex]
[tex]\[ f'(x) = \frac{1}{2\sqrt{x - 9}} \][/tex]
### Step 2: Evaluate the derivative at [tex]\( x = 13 \)[/tex]
To find the slope of the tangent line at the point [tex]\((13, 2)\)[/tex], substitute [tex]\( x = 13 \)[/tex] into the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(13) = \frac{1}{2\sqrt{13 - 9}} \][/tex]
[tex]\[ f'(13) = \frac{1}{2\sqrt{4}} \][/tex]
[tex]\[ f'(13) = \frac{1}{2 \cdot 2} \][/tex]
[tex]\[ f'(13) = \frac{1}{4} \][/tex]
Hence, the slope of the tangent line at [tex]\( x = 13 \)[/tex] is [tex]\( \frac{1}{4} \)[/tex].
### Step 3: Use the point-slope form of the equation of the line
The point-slope form of the equation of a line is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Here, [tex]\( (x_1, y_1) = (13, 2) \)[/tex] and the slope [tex]\( m = \frac{1}{4} \)[/tex]. Substituting these values, we get:
[tex]\[ y - 2 = \frac{1}{4}(x - 13) \][/tex]
### Step 4: Simplify the equation
To express the equation in the slope-intercept form [tex]\( y = mx + b \)[/tex], solve for [tex]\( y \)[/tex]:
[tex]\[ y - 2 = \frac{1}{4}(x - 13) \][/tex]
[tex]\[ y - 2 = \frac{1}{4}x - \frac{13}{4} \][/tex]
[tex]\[ y = \frac{1}{4}x - \frac{13}{4} + 2 \][/tex]
[tex]\[ y = \frac{1}{4}x - \frac{13}{4} + \frac{8}{4} \][/tex]
[tex]\[ y = \frac{1}{4}x - \frac{5}{4} \][/tex]
So, the equation of the tangent line to the graph of [tex]\( f(x) = \sqrt{x - 9} \)[/tex] at the point [tex]\((13, 2)\)[/tex] is:
[tex]\[ y = \frac{1}{4}x - \frac{5}{4} \][/tex]
### Summary
- The derivative of [tex]\( f(x) = \sqrt{x - 9} \)[/tex] is [tex]\( \frac{1}{2\sqrt{x - 9}} \)[/tex].
- The slope of the tangent line at [tex]\( x = 13 \)[/tex] is [tex]\( \frac{1}{4} \)[/tex].
- The equation of the tangent line at the point [tex]\((13, 2)\)[/tex] is [tex]\( y = \frac{1}{4}x - \frac{5}{4} \)[/tex].
### Step 1: Find the derivative of [tex]\( f(x) \)[/tex]
The function given is:
[tex]\[ f(x) = \sqrt{x - 9} \][/tex]
First, express the function as:
[tex]\[ f(x) = (x - 9)^{1/2} \][/tex]
To find the derivative of [tex]\( f(x) \)[/tex], use the power rule and the chain rule:
[tex]\[ f'(x) = \frac{1}{2}(x - 9)^{-1/2} \cdot (1) \][/tex]
[tex]\[ f'(x) = \frac{1}{2\sqrt{x - 9}} \][/tex]
### Step 2: Evaluate the derivative at [tex]\( x = 13 \)[/tex]
To find the slope of the tangent line at the point [tex]\((13, 2)\)[/tex], substitute [tex]\( x = 13 \)[/tex] into the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(13) = \frac{1}{2\sqrt{13 - 9}} \][/tex]
[tex]\[ f'(13) = \frac{1}{2\sqrt{4}} \][/tex]
[tex]\[ f'(13) = \frac{1}{2 \cdot 2} \][/tex]
[tex]\[ f'(13) = \frac{1}{4} \][/tex]
Hence, the slope of the tangent line at [tex]\( x = 13 \)[/tex] is [tex]\( \frac{1}{4} \)[/tex].
### Step 3: Use the point-slope form of the equation of the line
The point-slope form of the equation of a line is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Here, [tex]\( (x_1, y_1) = (13, 2) \)[/tex] and the slope [tex]\( m = \frac{1}{4} \)[/tex]. Substituting these values, we get:
[tex]\[ y - 2 = \frac{1}{4}(x - 13) \][/tex]
### Step 4: Simplify the equation
To express the equation in the slope-intercept form [tex]\( y = mx + b \)[/tex], solve for [tex]\( y \)[/tex]:
[tex]\[ y - 2 = \frac{1}{4}(x - 13) \][/tex]
[tex]\[ y - 2 = \frac{1}{4}x - \frac{13}{4} \][/tex]
[tex]\[ y = \frac{1}{4}x - \frac{13}{4} + 2 \][/tex]
[tex]\[ y = \frac{1}{4}x - \frac{13}{4} + \frac{8}{4} \][/tex]
[tex]\[ y = \frac{1}{4}x - \frac{5}{4} \][/tex]
So, the equation of the tangent line to the graph of [tex]\( f(x) = \sqrt{x - 9} \)[/tex] at the point [tex]\((13, 2)\)[/tex] is:
[tex]\[ y = \frac{1}{4}x - \frac{5}{4} \][/tex]
### Summary
- The derivative of [tex]\( f(x) = \sqrt{x - 9} \)[/tex] is [tex]\( \frac{1}{2\sqrt{x - 9}} \)[/tex].
- The slope of the tangent line at [tex]\( x = 13 \)[/tex] is [tex]\( \frac{1}{4} \)[/tex].
- The equation of the tangent line at the point [tex]\((13, 2)\)[/tex] is [tex]\( y = \frac{1}{4}x - \frac{5}{4} \)[/tex].
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