find an equation of the tangent line to the curve at y=2x^3+5x, (-1, 3)
dy/dx = 6x + 5
so at (-1,3), dy/dx = -6+5 = -1
slope = -1, point on line is (-1,3)
3 = (-1)(-1) + b in y = mx+b
2 = b
equation of tangent:
y = -x + 2
To find the equation of the tangent line to a curve at a given point, you need to find the derivative of the function representing the curve and then use the point-slope form of a line.
Step 1: Find the derivative of the function representing the curve.
Let's differentiate the function y = 2x^3 + 5x with respect to x using the power rule of differentiation:
dy/dx = 6x^2 + 5
Step 2: Use the point-slope form of a line to find the equation of the tangent line.
The point-slope form of a line is given by y - y1 = m(x - x1), where (x1, y1) is the given point on the curve.
Given that the point is (-1, 3), we can substitute these coordinates into the equation and the derivative we found:
y - 3 = (6x^2 + 5)(x - (-1))
Simplifying the equation, we have:
y - 3 = (6x^2 + 5)(x + 1)
Expanding the right side of the equation, we get:
y - 3 = 6x^3 + 11x^2 + 5x + 5
Finally, rearranging the equation into the standard form, we have:
y = 6x^3 + 11x^2 + 5x + 5 + 3
Simplifying the equation further, we get the equation of the tangent line:
y = 6x^3 + 11x^2 + 5x + 8