Find the point on the graph of y = x^2 where the tangent line is parallel to the line -5x+3y=5
To find the point on the graph where the tangent line is parallel to the line -5x + 3y = 5, we need to first determine the slope of the line.
We can rearrange the line equation into slope-intercept form (y = mx + b) by solving for y:
-5x + 3y = 5
3y = 5x + 5
y = (5/3)x + 5/3
The slope of the line is (5/3). Any tangent line to the graph of y = x^2 that is parallel to this line must have the same slope.
Now, let's find the derivative of the function y = x^2 to determine the slope of the tangent line at any given point on the graph:
dy/dx = 2x
To find the x-coordinate(s) where the slope of the tangent line is equal to (5/3), we can set the derivative equal to (5/3) and solve for x:
2x = 5/3
x = (5/3) / 2
x = 5/6
Therefore, at x = 5/6, the slope of the tangent line to the graph of y = x^2 will be parallel to the line -5x + 3y = 5.
To find the corresponding y-coordinate, substitute x = 5/6 back into the equation y = x^2:
y = (5/6)^2
y = 25/36
Therefore, the point on the graph of y = x^2 where the tangent line is parallel to the line -5x + 3y = 5 is (5/6, 25/36).