find all points at which the tangent line to the graph of y=x3 is parallel to the line y=27x+5
slope of the graph is 3x^2 for all x.
slope of the line is 27
so, for which x is 3x^2 = 27?
To find the points at which the tangent line to the graph of y = x^3 is parallel to the line y = 27x + 5, we need to determine the derivative of y = x^3 and set it equal to the derivative of y = 27x + 5.
Step 1: Find the derivative of y = x^3.
To do this, we use the power rule which states that the derivative of x^n (where n is a non-negative integer) is n*x^(n-1). In this case, n = 3.
So, dy/dx = 3*x^(3-1) = 3x^2.
Step 2: Set the derivative of y = x^3 equal to the derivative of y = 27x + 5.
Since we want the tangent line to be parallel to the line y = 27x + 5, the slopes of both lines should be equal. Therefore, set 3x^2 (the derivative of y = x^3) equal to 27 (the derivative of y = 27x + 5).
3x^2 = 27.
Step 3: Solve for x.
Divide both sides of the equation by 3:
x^2 = 9.
Taking the square root of both sides, we get:
x = ±3.
Step 4: Find the corresponding y-values.
Plug the values of x into the original equation y = x^3 to get the corresponding y-values.
For x = 3, y = 3^3 = 27.
For x = -3, y = (-3)^3 = -27.
So, the points at which the tangent line to the graph of y = x^3 is parallel to the line y = 27x + 5 are (3, 27) and (-3, -27).