Answer :
Sure! Let's break down the solution step by step.
### Step 1: Given Information
We are given the quadratic function in the form [tex]\( g(x) = ax^2 + bx + c \)[/tex], where:
[tex]\[ a = \frac{1}{2}, \quad b = -2, \quad c = 4 \][/tex]
### Step 2: Determine the Derivative
To find the equation of the tangent line, we need the derivative of the function [tex]\( g(x) \)[/tex]. The derivative of [tex]\( g(x) = ax^2 + bx + c \)[/tex] is:
[tex]\[ g'(x) = 2ax + b \][/tex]
### Step 3: Evaluate the Derivative at [tex]\( x = -2 \)[/tex]
To find the slope of the tangent line at [tex]\( x = -2 \)[/tex]:
[tex]\[ g'(-2) = 2 \left(\frac{1}{2}\right)(-2) + (-2) \][/tex]
[tex]\[ g'(-2) = 2 \cdot \frac{1}{2} \cdot (-2) - 2 \][/tex]
[tex]\[ g'(-2) = -2 - 2 \][/tex]
[tex]\[ g'(-2) = -4 \][/tex]
The slope of the tangent line at [tex]\( x = -2 \)[/tex] is [tex]\( -4 \)[/tex].
### Step 4: Determine the Point of Tangency at [tex]\( x = -2 \)[/tex]
To find the y-coordinate of the point of tangency, substitute [tex]\( x = -2 \)[/tex] into the original function [tex]\( g(x) \)[/tex]:
[tex]\[ g(-2) = \frac{1}{2}(-2)^2 - 2(-2) + 4 \][/tex]
[tex]\[ g(-2) = \frac{1}{2}(4) +4 + 4 \][/tex]
[tex]\[ g(-2) = 2 +4 + 4 \][/tex]
[tex]\[ g(-2) = 10 \][/tex]
So, the point of tangency is [tex]\((-2, 10)\)[/tex].
### Step 5: Equation of the Tangent Line
The equation of the line in point-slope form [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\((x_1, y_1)\)[/tex] is the point of tangency and [tex]\( m \)[/tex] is the slope:
[tex]\[ y - 10 = -4(x + 2) \][/tex]
### Step 6: Simplify the Equation
Distribute and rearrange the equation to get it into slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[ y - 10 = -4x - 8 \][/tex]
[tex]\[ y = -4x - 8 + 10 \][/tex]
[tex]\[ y = -4x + 2 \][/tex]
So, the equation of the tangent line at [tex]\( x = -2 \)[/tex] is:
[tex]\[ y = -4x + 2 \][/tex]
### Step 7: Tangent Line Parallel to the Given Tangent
For a line to be parallel, it must have the same slope. Therefore, the equation of the tangent to [tex]\( g(x) \)[/tex] that is parallel should also have a slope of [tex]\( -4 \)[/tex].
Using the same slope but different points, the general form for the equation of a line parallel to [tex]\( y = -4x + 2 \)[/tex]:
[tex]\[ y = -4x + k \][/tex]
where [tex]\( k \)[/tex] can be any constant determining the y-intercept of our new line.
Thus, any line of the form:
[tex]\[ y = -4x + k \][/tex]
where [tex]\( k \)[/tex] is a constant, will be parallel to the tangent line at [tex]\( x = -2 \)[/tex].
### Step 1: Given Information
We are given the quadratic function in the form [tex]\( g(x) = ax^2 + bx + c \)[/tex], where:
[tex]\[ a = \frac{1}{2}, \quad b = -2, \quad c = 4 \][/tex]
### Step 2: Determine the Derivative
To find the equation of the tangent line, we need the derivative of the function [tex]\( g(x) \)[/tex]. The derivative of [tex]\( g(x) = ax^2 + bx + c \)[/tex] is:
[tex]\[ g'(x) = 2ax + b \][/tex]
### Step 3: Evaluate the Derivative at [tex]\( x = -2 \)[/tex]
To find the slope of the tangent line at [tex]\( x = -2 \)[/tex]:
[tex]\[ g'(-2) = 2 \left(\frac{1}{2}\right)(-2) + (-2) \][/tex]
[tex]\[ g'(-2) = 2 \cdot \frac{1}{2} \cdot (-2) - 2 \][/tex]
[tex]\[ g'(-2) = -2 - 2 \][/tex]
[tex]\[ g'(-2) = -4 \][/tex]
The slope of the tangent line at [tex]\( x = -2 \)[/tex] is [tex]\( -4 \)[/tex].
### Step 4: Determine the Point of Tangency at [tex]\( x = -2 \)[/tex]
To find the y-coordinate of the point of tangency, substitute [tex]\( x = -2 \)[/tex] into the original function [tex]\( g(x) \)[/tex]:
[tex]\[ g(-2) = \frac{1}{2}(-2)^2 - 2(-2) + 4 \][/tex]
[tex]\[ g(-2) = \frac{1}{2}(4) +4 + 4 \][/tex]
[tex]\[ g(-2) = 2 +4 + 4 \][/tex]
[tex]\[ g(-2) = 10 \][/tex]
So, the point of tangency is [tex]\((-2, 10)\)[/tex].
### Step 5: Equation of the Tangent Line
The equation of the line in point-slope form [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\((x_1, y_1)\)[/tex] is the point of tangency and [tex]\( m \)[/tex] is the slope:
[tex]\[ y - 10 = -4(x + 2) \][/tex]
### Step 6: Simplify the Equation
Distribute and rearrange the equation to get it into slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[ y - 10 = -4x - 8 \][/tex]
[tex]\[ y = -4x - 8 + 10 \][/tex]
[tex]\[ y = -4x + 2 \][/tex]
So, the equation of the tangent line at [tex]\( x = -2 \)[/tex] is:
[tex]\[ y = -4x + 2 \][/tex]
### Step 7: Tangent Line Parallel to the Given Tangent
For a line to be parallel, it must have the same slope. Therefore, the equation of the tangent to [tex]\( g(x) \)[/tex] that is parallel should also have a slope of [tex]\( -4 \)[/tex].
Using the same slope but different points, the general form for the equation of a line parallel to [tex]\( y = -4x + 2 \)[/tex]:
[tex]\[ y = -4x + k \][/tex]
where [tex]\( k \)[/tex] can be any constant determining the y-intercept of our new line.
Thus, any line of the form:
[tex]\[ y = -4x + k \][/tex]
where [tex]\( k \)[/tex] is a constant, will be parallel to the tangent line at [tex]\( x = -2 \)[/tex].
