Determine the equation of a line that passes through (2,2) and is parallel to the line tangent to y=3x^3-2x at (-1,5)?
slope of y=3x^3-2x is 9x^2-2
so, slope at (-1,5) is 7
so, now yu have a point and a slope:
y-2 = 7(x-2)
Well, let's start by finding the derivative of the given function y = 3x^3 - 2x. Using the power rule, we get dy/dx = 9x^2 - 2.
Since the tangent line to the function at (-1,5) is parallel to the line we want to find, it must have the same slope. So, we can substitute x = -1 into the derivative to find the slope at that point.
Slope at (-1,5) = dy/dx |-1 = 9(-1)^2 - 2 = 9 - 2 = 7.
Now that we have the slope (m = 7) and a point the line passes through (2,2), we can use the point-slope formula to find the equation of the line.
y - y₁ = m(x - x₁)
y - 2 = 7(x - 2)
Expanding, we get:
y - 2 = 7x - 14
y = 7x - 12
So, the equation of the line parallel to the tangent line at (-1,5) and passing through (2,2) is y = 7x - 12.
To find the equation of a line that is parallel to the line tangent to y=3x^3-2x at (-1,5), we need to determine the slope of that tangent line.
First, we find the derivative of the function y=3x^3-2x using the power rule of derivatives:
dy/dx = 9x^2 - 2
Next, we evaluate the derivative at x = -1 to find the slope of the tangent line at (-1,5):
m = dy/dx |-1 = 9(-1)^2 - 2 = 9 - 2 = 7
So, the slope of the line tangent to y=3x^3-2x at (-1,5) is 7.
Since we want to find a line parallel to this tangent line, the slope of the new line will also be 7.
We already know that the new line passes through the point (2,2), so we can use the point-slope form of the equation of a line:
y - y1 = m(x - x1)
Plugging in m = 7, x1 = 2, and y1 = 2, we get:
y - 2 = 7(x - 2)
Expanding the equation:
y - 2 = 7x - 14
Simplifying:
y = 7x - 12
Therefore, the equation of the line that passes through (2,2) and is parallel to the line tangent to y=3x^3-2x at (-1,5) is y = 7x - 12.
To determine the equation of a line parallel to another line, we need to find the slope of the given line, and then use that slope to find the equation of the new line passing through the given point.
Step 1: Finding the slope of the tangent line
To find the slope of the tangent line to the curve y = 3x^3 - 2x at the point (-1, 5), we need to find the derivative of the given curve. Taking the derivative of the function, we have:
dy/dx = 9x^2 - 2
Evaluating this derivative at x = -1 (since we need the slope at that point), we get:
dy/dx = 9(-1)^2 - 2
= 9 - 2
= 7
Therefore, the slope of the tangent line at (-1, 5) is 7.
Step 2: Finding the equation of the line
Since the new line we want to find is parallel to the tangent line, it will have the same slope (i.e., 7). We can use the point-slope form of the line to find the equation. The point-slope form is given by:
y - y1 = m(x - x1)
Where (x1, y1) is a point on the line, and m is the slope of the line.
Using the given point (2, 2) and the slope 7, we can substitute these values into the point-slope form:
y - 2 = 7(x - 2)
Expanding and simplifying:
y - 2 = 7x - 14
Rearranging the equation to the slope-intercept form (y = mx + b):
y = 7x - 12
Therefore, the equation of the line passing through (2, 2) and parallel to the tangent line y = 3x^3 - 2x at (-1, 5) is y = 7x - 12.