Find the points on the graph of y = 1/x where the graph is parallel to the line 4x + 9y = 3.
Would the slope not be -4/9?
Oh, parallel lines, huh? Well, let me just step right up and help you out here! Now, if we want the graph of y = 1/x to be parallel to the line 4x + 9y = 3, we need the slopes of these two lines to be equal. So, let's get the slope of the line 4x + 9y = 3. Shall we? To find the slope, let's rewrite this equation in slope-intercept form.
4x + 9y = 3
9y = -4x + 3
y = (-4/9)x + 1/3
So, the slope of this line is -4/9. Now, looking at the equation y = 1/x, we can see that the slope of this line is 1. For these two lines to be parallel, their slopes must be equal. But wait a minute, -4/9 is NOT equal to 1! Ah, I guess the clowns messed up the trick this time! So, unfortunately, there are no points on the graph of y = 1/x that are parallel to the line 4x + 9y = 3. Guess we'll have to find something else to make the clowns at the circus happy!
To find the points on the graph of y = 1/x that are parallel to the line 4x + 9y = 3, we need to determine the slope of the line and match it with the slope of the graph of y = 1/x.
First, let's rearrange the equation 4x + 9y = 3 into the slope-intercept form y = mx + b, where m is the slope:
4x + 9y = 3
9y = -4x + 3
y = (-4/9)x + 3/9
y = (-4/9)x + 1/3
Comparing this equation with y = 1/x, we can see that the slope of the line 4x + 9y = 3 is -4/9.
To find points on the graph of y = 1/x that are parallel, we need to match the slopes. The slope of the graph of y = 1/x can be found by taking the derivative of the equation:
dy/dx = -1/x^2
To find where the slope is equal to -4/9, we set -1/x^2 equal to -4/9 and solve for x:
-1/x^2 = -4/9
Cross multiplying gives us:
-9 = -4x^2
Divide both sides of the equation by -4:
9/4 = x^2
Taking the square root of both sides gives us:
x = ±√(9/4) = ±(3/2)
Therefore, the points on the graph of y = 1/x that are parallel to the line 4x + 9y = 3 are (3/2, 2/3) and (-3/2, -2/3).
To find the points on the graph of y = 1/x where the graph is parallel to the line 4x + 9y = 3, we need to determine the slope of the given line.
The equation of a line in the form Ax + By = C can be written in slope-intercept form, y = -A/Bx + C/B. Comparing this with 4x + 9y = 3, we can see that its slope is -4/9.
Since we want to find points on the graph of y = 1/x that have the same slope, we can differentiate this equation implicitly. Differentiating y with respect to x, we get:
dy/dx = -1/x^2
To find the slope of the graph of y = 1/x at any given point, we substitute the x-coordinate of the point into dy/dx. Therefore, we need to find where dy/dx equals -4/9:
-1/x^2 = -4/9
Cross-multiplying, we get:
-9 = -4x^2
Dividing both sides by -4, we have:
9/4 = x^2
Taking the square root of both sides, we get:
x = ±sqrt(9/4) = ±3/2
So, the x-coordinates of the points where the graph of y = 1/x is parallel to the line 4x + 9y = 3 are x = -3/2 and x = 3/2.
Substituting these x-coordinates back into the equation y = 1/x, we can find the corresponding y-coordinates:
When x = -3/2, y = 1/(-3/2) = -2/3
When x = 3/2, y = 1/(3/2) = 2/3
Therefore, the points on the graph of y = 1/x that are parallel to the line 4x + 9y = 3 are (-3/2, -2/3) and (3/2, 2/3).
The slope of the straight line
4x + 9y = 3 is
-9/4.
You want points on the line y = 1/x where
dy/dx = -9/4. Therefore solve
-1/x^2 = -9/4
x^2 = 4/9
x = +2/3 and -2/3.
At those points, y = 3/2 and -3/2, respectively.
Therefore there are two points on the curve that satisfy the slope requirement:
(2/3, 3/2) and (-2/3, -3/2)