Find the x-coordinates of all points on the graph of y = x^3 - 4x^2 + 5x at which the tangent line is parallel to 2y - 10x - 7 = 0
the slope of the given line is 5
so find dy/dx for the cubic
and set that result equal to 5
you will get a quadratic which factors nicely.
let me know what values of x you got.
I got x=10/3 and x=6
3x^2 - 8x + 5 = 5
x(3x-8) = 0
x = 0 or x = 8/3
I don't understand how you got your answers.
To find the x-coordinates of the points on the graph of y = x^3 - 4x^2 + 5x at which the tangent line is parallel to 2y - 10x - 7 = 0, we need to determine the slope of the tangent line and find the values of x that satisfy that condition.
First, let's rewrite the equation of the tangent line in slope-intercept form:
2y - 10x - 7 = 0
2y = 10x + 7
y = 5x + 7/2
The slope of this line is 5, which means we need to find the points on the graph of y = x^3 - 4x^2 + 5x where the derivative of the function equals 5.
To find the derivative of the function y = x^3 - 4x^2 + 5x, we differentiate it with respect to x:
dy/dx = 3x^2 - 8x + 5
Next, we set the derivative equal to 5 and solve for x:
3x^2 - 8x + 5 = 5
3x^2 - 8x = 0
x(3x - 8) = 0
From this equation, we can see that either x = 0 or 3x - 8 = 0. Solving the second equation, we get:
3x - 8 = 0
3x = 8
x = 8/3
Therefore, the x-coordinates of the points on the graph of y = x^3 - 4x^2 + 5x where the tangent line is parallel to 2y - 10x - 7 = 0 are x = 0 and x = 8/3.