Answer :
To find the equations of the tangent and the normal to the curve at the point where [tex]\( x = 2 \)[/tex], we will proceed step-by-step.
The curve is given by the equation:
[tex]\[ y = \frac{8}{x} - x + 3x^2 \][/tex]
First, let's find the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] to determine the slope of the tangent line at the given point. The derivative of [tex]\( y \)[/tex] is:
[tex]\[ y' = -\frac{8}{x^2} - 1 + 6x \][/tex]
We need to evaluate this derivative at [tex]\( x = 2 \)[/tex] to get the slope of the tangent line at that point:
[tex]\[ y'(2) = -\frac{8}{2^2} - 1 + 6 \cdot 2 \][/tex]
[tex]\[ y'(2) = -\frac{8}{4} - 1 + 12 \][/tex]
[tex]\[ y'(2) = -2 - 1 + 12 \][/tex]
[tex]\[ y'(2) = 9 \][/tex]
So, the slope of the tangent line at [tex]\( x = 2 \)[/tex] is 9.
Next, we need to find the [tex]\( y \)[/tex]-coordinate of the curve at [tex]\( x = 2 \)[/tex]. We substitute [tex]\( x = 2 \)[/tex] into the original equation:
[tex]\[ y(2) = \frac{8}{2} - 2 + 3 \cdot 2^2 \][/tex]
[tex]\[ y(2) = 4 - 2 + 12 \][/tex]
[tex]\[ y(2) = 14 \][/tex]
So, the point of tangency is [tex]\((2, 14)\)[/tex].
Now, the equation of the tangent line in point-slope form is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\( m \)[/tex] is the slope of the tangent, and [tex]\((x_1, y_1)\)[/tex] is the point of tangency.
Substituting [tex]\( m = 9 \)[/tex], [tex]\( x_1 = 2 \)[/tex], and [tex]\( y_1 = 14 \)[/tex], we get:
[tex]\[ y - 14 = 9(x - 2) \][/tex]
[tex]\[ y - 14 = 9x - 18 \][/tex]
[tex]\[ y = 9x - 4 \][/tex]
So, the equation of the tangent line is:
[tex]\[ y = 9x - 4 \][/tex]
Next, we determine the equation of the normal line. The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is:
[tex]\[ m_{\text{normal}} = -\frac{1}{9} \][/tex]
Using the point-slope form again, the equation of the normal line at the point [tex]\((2, 14)\)[/tex] is:
[tex]\[ y - 14 = -\frac{1}{9}(x - 2) \][/tex]
[tex]\[ y - 14 = -\frac{1}{9}x + \frac{2}{9} \][/tex]
[tex]\[ y = -\frac{1}{9}x + \frac{2}{9} + 14 \][/tex]
[tex]\[ y = -\frac{1}{9}x + \frac{2}{9} + \frac{126}{9} \][/tex]
[tex]\[ y = -\frac{1}{9}x + \frac{128}{9} \][/tex]
So, the equation of the normal line is:
[tex]\[ y = -\frac{1}{9}x + \frac{128}{9} \][/tex]
In summary, the equations of the tangent and normal to the curve [tex]\( y = \frac{8}{x} - x + 3x^2 \)[/tex] at the point where [tex]\( x = 2 \)[/tex] are:
- Tangent line: [tex]\( y = 9x - 4 \)[/tex]
- Normal line: [tex]\( y = -\frac{1}{9}x + \frac{128}{9} \)[/tex]
The curve is given by the equation:
[tex]\[ y = \frac{8}{x} - x + 3x^2 \][/tex]
First, let's find the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] to determine the slope of the tangent line at the given point. The derivative of [tex]\( y \)[/tex] is:
[tex]\[ y' = -\frac{8}{x^2} - 1 + 6x \][/tex]
We need to evaluate this derivative at [tex]\( x = 2 \)[/tex] to get the slope of the tangent line at that point:
[tex]\[ y'(2) = -\frac{8}{2^2} - 1 + 6 \cdot 2 \][/tex]
[tex]\[ y'(2) = -\frac{8}{4} - 1 + 12 \][/tex]
[tex]\[ y'(2) = -2 - 1 + 12 \][/tex]
[tex]\[ y'(2) = 9 \][/tex]
So, the slope of the tangent line at [tex]\( x = 2 \)[/tex] is 9.
Next, we need to find the [tex]\( y \)[/tex]-coordinate of the curve at [tex]\( x = 2 \)[/tex]. We substitute [tex]\( x = 2 \)[/tex] into the original equation:
[tex]\[ y(2) = \frac{8}{2} - 2 + 3 \cdot 2^2 \][/tex]
[tex]\[ y(2) = 4 - 2 + 12 \][/tex]
[tex]\[ y(2) = 14 \][/tex]
So, the point of tangency is [tex]\((2, 14)\)[/tex].
Now, the equation of the tangent line in point-slope form is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\( m \)[/tex] is the slope of the tangent, and [tex]\((x_1, y_1)\)[/tex] is the point of tangency.
Substituting [tex]\( m = 9 \)[/tex], [tex]\( x_1 = 2 \)[/tex], and [tex]\( y_1 = 14 \)[/tex], we get:
[tex]\[ y - 14 = 9(x - 2) \][/tex]
[tex]\[ y - 14 = 9x - 18 \][/tex]
[tex]\[ y = 9x - 4 \][/tex]
So, the equation of the tangent line is:
[tex]\[ y = 9x - 4 \][/tex]
Next, we determine the equation of the normal line. The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is:
[tex]\[ m_{\text{normal}} = -\frac{1}{9} \][/tex]
Using the point-slope form again, the equation of the normal line at the point [tex]\((2, 14)\)[/tex] is:
[tex]\[ y - 14 = -\frac{1}{9}(x - 2) \][/tex]
[tex]\[ y - 14 = -\frac{1}{9}x + \frac{2}{9} \][/tex]
[tex]\[ y = -\frac{1}{9}x + \frac{2}{9} + 14 \][/tex]
[tex]\[ y = -\frac{1}{9}x + \frac{2}{9} + \frac{126}{9} \][/tex]
[tex]\[ y = -\frac{1}{9}x + \frac{128}{9} \][/tex]
So, the equation of the normal line is:
[tex]\[ y = -\frac{1}{9}x + \frac{128}{9} \][/tex]
In summary, the equations of the tangent and normal to the curve [tex]\( y = \frac{8}{x} - x + 3x^2 \)[/tex] at the point where [tex]\( x = 2 \)[/tex] are:
- Tangent line: [tex]\( y = 9x - 4 \)[/tex]
- Normal line: [tex]\( y = -\frac{1}{9}x + \frac{128}{9} \)[/tex]
