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Answer:
(0, 0), (6, 2)
Step-by-step explanation:
The derivative of a function gives us the slope of the tangent line at any given point on the graph.
Differentiate y = x/(x - 3) using the quotient rule.
[tex]\boxed{\begin{minipage}{5.5 cm}\underline{Quotient Rule for Differentiation}\\\\If $y=\dfrac{u}{v}$ then:\\\\\\$\dfrac{\text{d}y}{\text{d}x}=\dfrac{v \dfrac{\text{d}u}{\text{d}x}-u\dfrac{\text{d}v}{\text{d}x}}{v^2}$\\\end{minipage}}[/tex]
[tex]\textsf{Let}\;\;u=x \implies \dfrac{\text{d}u}{\text{d}x}=1[/tex]
[tex]\textsf{Let}\;\;v=x-3 \implies \dfrac{\text{d}v}{\text{d}x}=1[/tex]
Therefore:
[tex]\begin{aligned}\dfrac{\text{d}y}{\text{d}x}&=\dfrac{v \dfrac{\text{d}u}{\text{d}x}-u\dfrac{\text{d}v}{\text{d}x}}{v^2}\\\\&=\dfrac{(x-3)(1)-x(1)}{(x-3)^2}\\\\&=\dfrac{x-3-x}{(x-3)^2}\\\\&=-\dfrac{3}{(x-3)^2}\end{aligned}[/tex]
To find the points (x, y) on the graph of y = x/(x - 3) where the tangent lines are perpendicular to the line y = 3x - 8, we can use the fact that the slopes of perpendicular lines are negative reciprocals of each other.
The slope of the line y = 3x - 8 is 3.
Therefore, the slope of the perpendicular line is -1/3.
Set the derivative equal to the negative reciprocal -1/3 and solve for x:
[tex]\begin{aligned}\dfrac{\text{d}y}{\text{d}x}&=-\dfrac{1}{3}\\\\-\dfrac{3}{(x-3)^2}&=-\dfrac{1}{3}\\\\\dfrac{3}{(x-3)^2}&=\dfrac{1}{3}\\\\3 \cdot 3&=1 \cdot (x-3)^2\\\\9&=(x-3)^2\\\\(x-3)^2&=9\\\\x-3&=\pm3\\\\x&=3\pm3\\\\\implies x&=0\\\implies x&=6\end{aligned}[/tex]
Substitute the found values of x into the original equation y = x/(x - 3) to find the corresponding y-values:
[tex]\begin{aligned}x=0 \implies y&=\dfrac{0}{0-3}\\\\&=\dfrac{0}{-3}\\\\&=0\end{aligned}[/tex] [tex]\begin{aligned}x=6 \implies y&=\dfrac{6}{6-3}\\\\&=\dfrac{6}{3}\\\\&=2\end{aligned}[/tex]
Therefore, the points (x, y) on the graph of y = x/(x - 3) with tangent lines perpendicular to y = 3x - 8 are (0, 0) and (6, 2).
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Rewritten by : Barada
The points (x,y) on the graph of y = x/(x - 3) with tangent lines perpendicular to the line y = 3x - 8 are
The tangent to a curve is the derivative of the curve at that point.
To find all points (x,y) on the graph of y = x/(x - 3) with tangent lines perpendicular to the line y = 3x - 8
We proceed as follows
Since y = x/(x - 3), to find the tangent to the graph, we find the derivative of the graph.
Using the product rule dy/dx = (vdu/dx - udv/dx)/v² where
So, substituting the values of the variables into the equation, we have that
dy/dx = (vdu/dx - udv/dx)/v²
dy/dx = [(x - 3) × 1 - x × 1)/(x - 3)²
= [(x - 3) - x]/(x - 3)²
= [x - 3 - x]/(x - 3)²
= [x - x - 3]/(x - 3)²
= (0 - 3)/(x - 3)²
= - 3/(x - 3)²
Now, the line y = 3x - 8 has the gradient m = 3. For the tangent lines with gradient m', to be perpendicular to this line, we know that
mm' = -1 (for perpendicular lines)
So, m' = -1/m
= -1/3
Since m' = dy/dx, we have that
- 3/(x - 3)² = -1/3
Cross-multiplying, we have that
(x - 3)² = - 3 × - 3
(x - 3)² = 9
Taking square root of both sides, we have that
x - 3 = ±√9
x - 3 = ±3
x = ±3 + 3
x = -3 + 3 or x = 3 + 3
x = 0 or x = 6
Substituting these into y, we have that
y = x(x - 3)
When x = 0, y = 0/(0 -3) = 0/-3 =0
When x = 6, y = 6/(6 - 3) = 6/3 = 2
So, the points are (0, 0) and (6, 2)
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