This is from our problem set.

f(x)= 2(x^3)-3(x^2)-12x+20

write an equation of the line perpendicular to the graph of f at x=0

I think i'm supposed to take the derivate of f and then find the slope of the tangent line at x=0. Then take the opposite reciporcal of that and put this slope and the point (0,20) into point-slope from to get the equation but i'm not exactly sure if this is write. So, can someone help me.

Thank YOu

You are on the right track! To find the equation of a line perpendicular to the graph of f at x=0, you can follow these steps:

1. Take the derivative of f(x) to find the slope of the tangent line at x=0.
2. Find the negative reciprocal of the slope obtained in step 1 to get the slope of the perpendicular line.
3. Use the point-slope form of a linear equation, y - y₁ = m(x - x₁), where (x₁, y₁) is the given point (0, 20) and m is the slope from step 2.
4. Substitute the values into the equation in step 3 to get the equation of the line perpendicular to f at x=0.

Now let's go through the steps in detail:

Step 1: Find the derivative of f(x)
The derivative of f(x) = 2x^3 - 3x^2 - 12x + 20 can be found by using the power rule:
f'(x) = (2 * 3 x^(3-1)) - (3 * 2 x^(2-1)) - (12) = 6x^2 - 6x - 12

Step 2: Find the slope of the perpendicular line
The slope of the tangent line at x=0 is given by evaluating f'(x) at x=0:
f'(0) = 6(0)^2 - 6(0) - 12 = -12

Since you want the slope of the perpendicular line, take the negative reciprocal of -12:
Slope of perpendicular line = -(1 / -12) = 1/12

Step 3: Use the point-slope form
Now, use the point-slope form of a linear equation, y - y₁ = m(x - x₁), where (x₁, y₁) = (0, 20) and m = 1/12:
y - 20 = (1/12)(x - 0)

Step 4: Simplify the equation
Simplify the equation from step 3:
y - 20 = (1/12)x
Rearrange the equation to get it in slope-intercept form (y = mx + b):
y = (1/12)x + 20

So, the equation of the line perpendicular to the graph of f at x=0 is y = (1/12)x + 20.

I hope this explanation helps! Let me know if you have any further questions.