To find the points on the curve where the tangent is horizontal, we need to start by analyzing the derivative of the function. A horizontal tangent occurs when the derivative of the function is equal to zero.
Assuming the function is given as:
y = 2x^3 + 3x^2 + 12x^2
First, let’s simplify this equation:
y = 2x^3 + 15x^2 (since 3x² + 12x² = 15x²).
Next, we compute the derivative of y with respect to x:
y’ = d/dx(2x^3 + 15x^2) = 6x^2 + 30x.
To find where the tangent is horizontal, we set the derivative equal to zero:
6x^2 + 30x = 0.
Factoring out the common terms gives:
6x(x + 5) = 0.
This equation yields two solutions:
x = 0 and x = -5.
Now, we need to find the corresponding y values for these x values by substituting them back into the original equation:
1. For x = 0:
y = 2(0)^3 + 15(0)^2 = 0.
So the point is (0, 0).
2. For x = -5:
y = 2(-5)^3 + 15(-5)^2.
y = 2(-125) + 15(25) = -250 + 375 = 125.
So the point is (-5, 125).
In conclusion, the points on the curve where the tangent is horizontal are:
- (0, 0)
- (-5, 125)