Answer :
To find the equation of the tangent line to the function [tex]\( f(x) = x^2 - 3x + 5 \)[/tex] at the point [tex]\( (2, 3) \)[/tex], follow these steps:
### Step 1: Find the derivative of [tex]\( f(x) \)[/tex]
The derivative [tex]\( f'(x) \)[/tex] represents the slope of the tangent line at any point [tex]\( x \)[/tex].
Given the function:
[tex]\[ f(x) = x^2 - 3x + 5 \][/tex]
Its derivative is:
[tex]\[ f'(x) = \frac{d}{dx}(x^2 - 3x + 5) \][/tex]
[tex]\[ f'(x) = 2x - 3 \][/tex]
### Step 2: Evaluate the derivative at [tex]\( x = 2 \)[/tex]
To determine the slope of the tangent line at [tex]\( x = 2 \)[/tex]:
[tex]\[ f'(2) = 2(2) - 3 \][/tex]
[tex]\[ f'(2) = 4 - 3 \][/tex]
[tex]\[ f'(2) = 1 \][/tex]
So, the slope of the tangent line at the point [tex]\( (2, 3) \)[/tex] is 1.
### Step 3: Use the point-slope form to find the equation of the tangent line
The point-slope form of the equation of a line is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\( (x_1, y_1) \)[/tex] is the given point and [tex]\( m \)[/tex] is the slope.
Given:
- Slope ([tex]\( m \)[/tex]) = 1
- Point ([tex]\( x_1, y_1 \)[/tex]) = (2, 3)
Substituting into the point-slope form:
[tex]\[ y - 3 = 1(x - 2) \][/tex]
### Step 4: Simplify the equation of the tangent line
Solve for [tex]\( y \)[/tex] to put the equation in slope-intercept form ( [tex]\( y = mx + b \)[/tex] ):
[tex]\[ y - 3 = x - 2 \][/tex]
[tex]\[ y = x - 2 + 3 \][/tex]
[tex]\[ y = x + 1 \][/tex]
Therefore, the equation of the tangent line passing through the point [tex]\( (2, 3) \)[/tex] of the function [tex]\( f(x) = x^2 - 3x + 5 \)[/tex] is:
[tex]\[ \boxed{y = x + 1} \][/tex]
Hence, the correct answer is:
d. [tex]\( y = x + 1 \)[/tex]
### Step 1: Find the derivative of [tex]\( f(x) \)[/tex]
The derivative [tex]\( f'(x) \)[/tex] represents the slope of the tangent line at any point [tex]\( x \)[/tex].
Given the function:
[tex]\[ f(x) = x^2 - 3x + 5 \][/tex]
Its derivative is:
[tex]\[ f'(x) = \frac{d}{dx}(x^2 - 3x + 5) \][/tex]
[tex]\[ f'(x) = 2x - 3 \][/tex]
### Step 2: Evaluate the derivative at [tex]\( x = 2 \)[/tex]
To determine the slope of the tangent line at [tex]\( x = 2 \)[/tex]:
[tex]\[ f'(2) = 2(2) - 3 \][/tex]
[tex]\[ f'(2) = 4 - 3 \][/tex]
[tex]\[ f'(2) = 1 \][/tex]
So, the slope of the tangent line at the point [tex]\( (2, 3) \)[/tex] is 1.
### Step 3: Use the point-slope form to find the equation of the tangent line
The point-slope form of the equation of a line is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\( (x_1, y_1) \)[/tex] is the given point and [tex]\( m \)[/tex] is the slope.
Given:
- Slope ([tex]\( m \)[/tex]) = 1
- Point ([tex]\( x_1, y_1 \)[/tex]) = (2, 3)
Substituting into the point-slope form:
[tex]\[ y - 3 = 1(x - 2) \][/tex]
### Step 4: Simplify the equation of the tangent line
Solve for [tex]\( y \)[/tex] to put the equation in slope-intercept form ( [tex]\( y = mx + b \)[/tex] ):
[tex]\[ y - 3 = x - 2 \][/tex]
[tex]\[ y = x - 2 + 3 \][/tex]
[tex]\[ y = x + 1 \][/tex]
Therefore, the equation of the tangent line passing through the point [tex]\( (2, 3) \)[/tex] of the function [tex]\( f(x) = x^2 - 3x + 5 \)[/tex] is:
[tex]\[ \boxed{y = x + 1} \][/tex]
Hence, the correct answer is:
d. [tex]\( y = x + 1 \)[/tex]
