Write the equation of the tangent line for the curve y=x^3+3x-8 at point (2,6)
Would I take the derivative of the equation and then plug in 2 and then plug everything into y-y1=m(x-x1)?
yes
Would the equation be y-6=15(x-2)?
Yes, you are correct. To find the equation of the tangent line at a given point on a curve, you can follow these steps:
1. Take the derivative of the equation to find the slope of the tangent line at any point on the curve.
2. Plug in the x-coordinate of the given point into the derivative equation to find the slope at that point.
3. Use the slope and the given point (x1, y1) to find the equation of the tangent line using the point-slope form, which is y - y1 = m(x - x1).
Let's go through the steps and find the equation of the tangent line for the curve y = x^3 + 3x - 8 at the point (2, 6).
Step 1: Take the derivative of the equation:
The derivative of y = x^3 + 3x - 8 is dy/dx = 3x^2 + 3.
Step 2: Plug in x = 2 into the derivative equation to find the slope at that point:
dy/dx = 3(2)^2 + 3 = 15.
Step 3: Use the slope (m = 15) and the point (2, 6) to find the equation of the tangent line:
Using the point-slope form y - y1 = m(x - x1), we substitute the values:
y - 6 = 15(x - 2).
Simplifying the equation:
y - 6 = 15x - 30.
y = 15x - 24.
So, the equation of the tangent line for the curve y = x^3 + 3x - 8 at the point (2, 6) is y = 15x - 24.
Yes, you are on the right track. To find the equation of the tangent line for the curve y = x^3 + 3x - 8 at the point (2, 6), you need to follow these steps:
Step 1: Take the derivative of the equation.
To find the derivative of y = x^3 + 3x - 8, you can apply the power rule, which states that for any term of the form x^n, the derivative is nx^(n-1).
So, differentiating y = x^3 + 3x - 8 gives us dy/dx = 3x^2 + 3.
Step 2: Plug in the x-coordinate (2) into the derivative.
To find the slope of the tangent line at the point (2, 6), substitute x = 2 into the derivative equation we obtained in Step 1:
dy/dx = 3(2)^2 + 3 = 15.
Therefore, the slope of the tangent line at (2, 6) is 15.
Step 3: Use the point-slope form to write the equation of the tangent line.
Using the point-slope form, which states that y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is a point on the line, we can substitute the known values into this equation.
In this case, the known values are: m = 15, x₁ = 2, and y₁ = 6.
Plugging them into the equation, we get:
y - 6 = 15(x - 2).
This is the equation of the tangent line for the given curve at the point (2, 6).