function f(x)=3x^2-2x-1 find equation of the line that is tangent and parallel to 8x-2y+12=0
To find the equation of the line that is parallel to another line and tangent to a given function, we can follow these steps:
Step 1: Determine the slope (m) of the given line. It can be obtained by rearranging the equation into the slope-intercept form (y = mx + c) and identifying the coefficient of x. In this case, the given line is 8x - 2y + 12 = 0. Rearranging, we get:
2y = 8x + 12
y = 4x + 6
Comparing with y = mx + c, we see that the slope (m) is 4.
Step 2: Find the derivative of the function f(x) with respect to x. The derivative measures the rate at which the function is changing at each point and gives us the slope of the tangent line. In this case, f(x) = 3x^2 - 2x - 1. Taking the derivative, we have:
f'(x) = 6x - 2
Step 3: Set the derivative equal to the slope of the given line to find the x-coordinate(s) where the tangent line is parallel. Since parallel lines have the same slope, we equate the derivative f'(x) to the slope of the given line: 6x - 2 = 4. Solving this equation gives:
6x - 2 = 4
6x = 6
x = 1
Step 4: Substitute the x-coordinate(s) obtained from Step 3 into the original function f(x) to find the corresponding y-coordinate(s). In this case, when x = 1, we have:
f(1) = 3(1)^2 - 2(1) - 1
f(1) = 0
Step 5: Use the obtained point (x, y) and the slope (m) of the given line to find the equation of the tangent line using the point-slope form (y - y1 = m(x - x1)). Substituting the values, we have:
y - y1 = m(x - x1)
y - 0 = 4(x - 1)
y = 4x - 4
Therefore, the equation of the line that is both parallel to the line 8x - 2y + 12 = 0 and tangent to the function f(x) = 3x^2 - 2x - 1 is y = 4x - 4.