The slope of the tangent line to the curve y= 3x^3 at the point (-1,-3) is:?
The equation of this tangent line can be written in the form y=mx+b where m is:?
and where b is:?
dy/dx = 9x^2
at (-1,-3) , dy/dx = 9(1) = 9
equation of tangent
y = 9x + b
but (-1,-3) lies on it,
-3 = 9(-1) + b
6 = b
y = 9x + 6 , m = 9 and b = 6
Well, the slope of the tangent line can be found by taking the derivative of the function and evaluating it at the given point. So first, let's find the derivative of y = 3x^3:
dy/dx = 9x^2
Now let's find the slope by plugging in x = -1:
dy/dx = 9(-1)^2
dy/dx = 9(1)
dy/dx = 9
So, the slope (m) of the tangent line is 9.
To find the y-intercept (b), we can use the point (-1, -3) and the slope (m = 9) in the equation of a line y = mx + b:
-3 = 9(-1) + b
-3 = -9 + b
b = -3 + 9
b = 6
Therefore, the equation of the tangent line is y = 9x + 6, where the slope (m) is 9 and the y-intercept (b) is 6.
To find the slope of the tangent line to the curve y = 3x^3 at the point (-1, -3), we need to find the derivative of the function.
Step 1: Find the derivative of y = 3x^3.
Taking the derivative of a polynomial involves multiplying each term by its power and decreasing the power by 1.
The derivative of 3x^3 is (3 * 3)x^(3-1), which simplifies to 9x^2.
Step 2: Evaluate the derivative at x = -1.
Plug x = -1 into the derivative function to find the slope at that point.
m = 9(-1)^2
m = 9(1)
m = 9
Therefore, the slope of the tangent line to the curve y = 3x^3 at the point (-1, -3) is 9.
To find the equation of the tangent line in the form y = mx + b, we need the slope (m) and the coordinates of the point (-1, -3).
Step 3: Substitute the slope and the coordinates of the point into the equation y = mx + b.
Using the point-slope form, we have:
-3 = 9(-1) + b
Step 4: Simplify and solve for b.
-3 = -9 + b
b = -3 + 9
b = 6
Therefore, the equation of the tangent line can be written as y = 9x + 6, where m = 9 and b = 6.
To find the slope of the tangent line to the curve y = 3x^3 at the point (-1, -3), we need to find the derivative of the function and evaluate it at x = -1.
Step 1: Find the derivative of the function y = 3x^3.
To find the derivative, we need to use the power rule for differentiation. The power rule states that if we have a term of the form x^n, then the derivative is nx^(n-1).
Applying the power rule, the derivative of y = 3x^3 is:
dy/dx = 3 * 3x^(3-1) = 9x^2
Step 2: Evaluate the derivative at x = -1.
Substitute x = -1 into the derivative function dy/dx.
dy/dx = 9(-1)^2 = 9(1) = 9
The slope of the tangent line to the curve at the point (-1, -3) is 9.
To find the equation of the tangent line in the form y = mx + b, we need to find the y-intercept (b) by substituting the point (-1, -3) and the slope (m = 9) into the equation.
Using the point-slope form of a line (y - y1 = m(x - x1)), we can write the equation:
y - (-3) = 9(x - (-1))
y + 3 = 9(x + 1)
Simplifying the equation, we get:
y + 3 = 9x + 9
Rearranging the equation, we get it in the form y = mx + b:
y = 9x + 6
Therefore, the equation of the tangent line is y = 9x + 6, where m = 9 is the slope and b = 6 is the y-intercept.