find an equation for the line tangent to y=2-6x^2 at (-5,-148)
dy/dx = -12x
so at (-5,-148) , slope = 60
y + 148 = 60(x+5)
60x - y = -152
14(-1)-27
To find the equation of the tangent line to the curve y = 2 - 6x^2 at the point (-5, -148), we can use the concept of differentiation. The derivative of a function represents the rate of change of the function at any given point.
Step 1: Find the derivative of the given function.
The derivative of y = 2 - 6x^2 can be found by applying the power rule for differentiation. Since the power rule states that d/dx (x^n) = n*x^(n-1), the derivative of -6x^2 is:
dy/dx = -12x
Step 2: Substitute the x-coordinate of the given point (-5, -148) into the derivative dy/dx to find the slope of the tangent line.
Substituting x = -5 into dy/dx = -12x:
dy/dx = -12 * (-5) = 60
Step 3: Use the slope-intercept form of a linear equation to find the equation of the tangent line.
The slope-intercept form of a linear equation is y = mx + b, where m is the slope of the line, and b is the y-intercept.
Since we know the slope of the tangent line is 60 and that it passes through the point (-5, -148), we can substitute these values into the slope-intercept form:
y = mx + b
-148 = 60*(-5) + b
Simplifying the equation gives:
-148 = -300 + b
Adding 300 to both sides:
b = -148 + 300
b = 152
Therefore, the equation of the tangent line is:
y = 60x + 152.