Minimum slope of a line tangent to the curve HELP PLEASE

Question

Find the minimum slope of the line tangent to the curve y = x³-3x²+2x+9

Answer 1

Hi,

If you plot this function you will see that it has two places where the slope is zero.

We can find those values of x where this is true by setting the derivative (the slope) equal to 0.

y(x)=x³-3x²+2x+9

dy/dx=3x²-6x+2

Now if we set dy/dx=0 we have a quadratic equation to solve

3x²-6x+2=0 find the roots and you will have found which values of x will locate the minimum slope of y(x)

Let me know if you have ??

Jim

Answer 2

By definition. the slope is the 1st derivative of a function.

SO:

y = x³ - 3x² + 2x + 9

y' = 3x² -6x + 2

Then, the minimum slope is when the 2nd derivative is zero:

y" = 6x - 6 = 0

x = 1

Thus, the minimum slope is 3*1² - 6*1 + 2 = 3 - 6 + 2 = -1

I am assuming that "minimum" means most negative. To find the minimum absolute value, set the 1st derivative = 0. This gives us x = 1 ± (√3 / 3)

Plot the curves to see how this works---use an online plotting tool such as this:

https://www.desmos.com/calculator

Answer 3

Whenever you hear "slope of the tangent line", think derivative. The derivative of y = x³-3x²+2x+9 is

y' = 3x² - 6x + 2

The problem asks to find the minimum value of y'.There are several ways to do this. The quickest way is to recognize that f(x) = 3x² - 6x + 2 is the equation of a parabola that opens upward. Its minimum is at the vertex (-^b/_2a, f(-^b/_2a)) where a = 3 and b = -6:

-^b/_2a = - ^(-6)/₂₍₃₎ = 1

f(-^b/_2a) = 3(1)² - 6(1) + 2 = -1

Therefore, the minimum slope of y = x³-3x²+2x+9 is y' = -1, which occurs at x = 1.