3.Given the function f defined by f(x)=2x^3-3x^2-12x+20

a.Find the zeros of f
b.Write an equation of the line perpendicular to the graph of f at x = 0
c. Find the x and y coordinates of all points on the graph of f where the line tangent to the graph is parallel to the x axis.

For (a), do some trial solutions by substituting x=factors of 20, such as

f(1),f(2),f(5),f(10), etc.
If you find one (there is one distinct root, and two coincident roots). The coincident roots are positive (as can be guessed by the Descartes rule of signs).

Once you found one, you just have to divide f(x) by the first zero to get a quadratic, which can then be factored by standard methods.

If you need further help, please post your attempts.

I'm in AP calc. We still didn't learn what descartes rule is.. Can you be more explisive please?

Descartes rule of signs is probably not important in this context,but here it is:

If the polynomial is arranged in descending order of the powers of the (single) variable, the number of change of signs indicates the maximum number of positive zeroes.
In the given case, there are two changes of sign, so the number of positive roots is either two or zero.
(see following link for a more detailed version)
http://en.wikipedia.org/wiki/Descartes%27_rule_of_signs

In solving for rational roots of a cubic, one of the ways is to try all possible combinations of the following as zeroes:
(factors of the constant term)/(factors of the leading coefficient).

In the case in point, we need to try as zeroes all of:
(±20,±10,±5,±2,±1)/(±2,±1).

At the beginning it looks like it's a lot of work, but as you get more experienced, it will be very straightforward. You just have to get organized.

For example, check the values of (start with small values, they are usually more likely to be zeroes):
f(1/2)
f(-1/2)
f(1)
f(-1)
f(2)
f(-2)
f(5)
f(-5)
f(5/2)
f(-5/2)
f(10)
f(-10)
f(20)
f(-20)
These are all the possible rational roots to try.

a. To find the zeros of a function, such as f(x), we need to set f(x) equal to zero and solve for x. In this case, we have the equation:

2x^3 - 3x^2 - 12x + 20 = 0

To solve this equation, we can use various methods such as factoring, the quadratic formula, or numerical methods like graphing. However, in this case, factoring and the quadratic formula won't work as the equation is a cubic equation.

One way to find solutions to this cubic equation is by using a numerical method like graphing. By plotting the graph of the function, we can visually determine the values of x where f(x) crosses the x-axis, indicating the zeros of the function.

Alternatively, we can use a graphing calculator or a computer algebra system to find the zeros of the function. These tools are typically equipped with algorithms that can find the zeros of a function accurately.

b. To find the equation of a line perpendicular to the graph of f at x = 0, we need to first find the slope of the graph of f at that point. The slope of the tangent line at a given point on the graph of a function is equal to the derivative of the function at that point.

To find the derivative of f(x), we can use the power rule, which states that for a function of the form f(x) = ax^n, the derivative is given by f'(x) = nax^(n-1).

Applying the power rule, we find that the derivative of f(x) = 2x^3 - 3x^2 - 12x + 20 is:

f'(x) = 6x^2 - 6x - 12

Now, we can evaluate the derivative at x = 0 to find the slope of the tangent line at that point:

f'(0) = 6(0)^2 - 6(0) - 12
= -12

Since we want the equation of a line perpendicular to the graph of f at x = 0, the slope of this line will be the negative reciprocal of -12, which is 1/12.

Therefore, the equation of the line perpendicular to the graph of f at x = 0 can be written in the form y = mx + b, where m is the slope and b is the y-intercept. Since the line is perpendicular to the x-axis at x = 0, the line will intersect the y-axis at y = 0.

Hence, the equation of the line is y = (1/12)x.

c. To find the x and y coordinates of all points on the graph of f where the tangent line is parallel to the x-axis, we need to find the points where the derivative of f(x) is equal to zero. These points correspond to the extrema (maximum or minimum) of the function.

The derivative of f(x) is given by:

f'(x) = 6x^2 - 6x - 12

To find the x-coordinates where the tangent line is parallel to the x-axis, we need to solve the equation:

6x^2 - 6x - 12 = 0

We can use various methods to solve this quadratic equation like factoring, completing the square, or using the quadratic formula. In this case, using the quadratic formula is the most convenient method:

x = (-(-6) ± √((-6)^2 - 4(6)(-12))) / (2*6)
x = (6 ± √(36 + 288)) / 12
x = (6 ± √(324)) / 12
x = (6 ± 18) / 12

Thus, the two x-coordinates where the tangent line is parallel to the x-axis are:

x = (6 + 18) / 12 = 24 / 12 = 2
x = (6 - 18) / 12 = -12 / 12 = -1

To find the corresponding y-coordinates, we substitute these x-values into the original function f(x):

For x = 2:
f(2) = 2(2)^3 - 3(2)^2 - 12(2) + 20
= 16 - 12 - 24 + 20
= 0

Therefore, the point (2, 0) is on the graph of f where the tangent line is parallel to the x-axis.

For x = -1:
f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 20
= -2 - 3 + 12 + 20
= 27

Thus, the point (-1, 27) is on the graph of f where the tangent line is parallel to the x-axis.

In summary, the x and y coordinates of all points on the graph of f where the tangent line is parallel to the x-axis are (2, 0) and (-1, 27).