Answer :
Of course! Let's go through the solution step-by-step.
### 1. Finding the slope [tex]\( m \)[/tex] of the tangent line at the point [tex]\((4, 5)\)[/tex]
Given the function:
[tex]\[ y = 3x^2 - 11x + 1 \][/tex]
To find the slope of the tangent line at a specific point, we need to calculate the derivative of the function with respect to [tex]\( x \)[/tex].
The derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}(3x^2 - 11x + 1) \][/tex]
Using standard derivative rules:
[tex]\[ \frac{dy}{dx} = 6x - 11 \][/tex]
Next, we evaluate this derivative at the given point [tex]\( x = 4 \)[/tex]:
[tex]\[ m = 6(4) - 11 \][/tex]
[tex]\[ m = 24 - 11 \][/tex]
[tex]\[ m = 13 \][/tex]
So, the slope [tex]\( m \)[/tex] of the tangent line at the point [tex]\((4, 5)\)[/tex] is 13.
### 2. Finding the equation of the tangent line at the point [tex]\((4, 5)\)[/tex]
The equation of a line in point-slope form is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Here, [tex]\((x_1, y_1)\)[/tex] is the point [tex]\((4, 5)\)[/tex], and [tex]\( m = 13 \)[/tex]. Plugging in these values, we get:
[tex]\[ y - 5 = 13(x - 4) \][/tex]
To find the equation in slope-intercept form ([tex]\( y = mx + b \)[/tex]), we'll simplify the above equation:
[tex]\[ y - 5 = 13x - 52 \][/tex]
Adding 5 to both sides:
[tex]\[ y = 13x - 52 + 5 \][/tex]
[tex]\[ y = 13x - 47 \][/tex]
So, the equation of the tangent line at the point [tex]\((4, 5)\)[/tex] is:
[tex]\[ y = 13x - 47 \][/tex]
### Summary
- The slope of the tangent line at the point [tex]\((4, 5)\)[/tex] is [tex]\( m = 13 \)[/tex].
- The equation of the tangent line at the point [tex]\((4, 5)\)[/tex] is [tex]\( y = 13x - 47 \)[/tex].
### 1. Finding the slope [tex]\( m \)[/tex] of the tangent line at the point [tex]\((4, 5)\)[/tex]
Given the function:
[tex]\[ y = 3x^2 - 11x + 1 \][/tex]
To find the slope of the tangent line at a specific point, we need to calculate the derivative of the function with respect to [tex]\( x \)[/tex].
The derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}(3x^2 - 11x + 1) \][/tex]
Using standard derivative rules:
[tex]\[ \frac{dy}{dx} = 6x - 11 \][/tex]
Next, we evaluate this derivative at the given point [tex]\( x = 4 \)[/tex]:
[tex]\[ m = 6(4) - 11 \][/tex]
[tex]\[ m = 24 - 11 \][/tex]
[tex]\[ m = 13 \][/tex]
So, the slope [tex]\( m \)[/tex] of the tangent line at the point [tex]\((4, 5)\)[/tex] is 13.
### 2. Finding the equation of the tangent line at the point [tex]\((4, 5)\)[/tex]
The equation of a line in point-slope form is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Here, [tex]\((x_1, y_1)\)[/tex] is the point [tex]\((4, 5)\)[/tex], and [tex]\( m = 13 \)[/tex]. Plugging in these values, we get:
[tex]\[ y - 5 = 13(x - 4) \][/tex]
To find the equation in slope-intercept form ([tex]\( y = mx + b \)[/tex]), we'll simplify the above equation:
[tex]\[ y - 5 = 13x - 52 \][/tex]
Adding 5 to both sides:
[tex]\[ y = 13x - 52 + 5 \][/tex]
[tex]\[ y = 13x - 47 \][/tex]
So, the equation of the tangent line at the point [tex]\((4, 5)\)[/tex] is:
[tex]\[ y = 13x - 47 \][/tex]
### Summary
- The slope of the tangent line at the point [tex]\((4, 5)\)[/tex] is [tex]\( m = 13 \)[/tex].
- The equation of the tangent line at the point [tex]\((4, 5)\)[/tex] is [tex]\( y = 13x - 47 \)[/tex].
