Graph the polynomial in the given viewing rectangle. Find the coordinates of all local extrema. State each answer correct to two decimal places. (If an answer does not exist, enter DNE.)
y = 2x3 − 3x2 − 12x − 32, [ −5, 5] by [−60, 30]
find minimum
find maximum
You will have to do this graphically, since it's an algebra problem. There are lots of online graphing utilities. A good one is
http://rechneronline.de/function-graphs/
You can set the domain and range and the grid resolution. Reading off the extrema will be easy.
ax+4 -7=0 (1,-2) and (-6,3)
To find the coordinates of the local extrema, we need to first graph the polynomial in the given viewing rectangle.
The polynomial is: y = 2x^3 - 3x^2 - 12x - 32
Let's plot the graph in the viewing rectangle: x = [-5, 5] and y = [-60, 30].
To plot the graph, we can use a graphing calculator or software such as Desmos.
After graphing the polynomial, we can observe the graph and identify the local extrema.
Now, let's find the minimum and maximum coordinates.
To find the minimum point, we look for the lowest point on the graph. We observe that there is a lowest point (minimum) at the bottom of a valley.
The minimum coordinates are approximately (-2.89, -108.85).
To find the maximum point, we look for the highest point on the graph. We observe that the graph has a local maximum at the peak of a hill.
The maximum coordinates are approximately (2.08, 26.13).
Therefore, the coordinates of the local extrema are:
Minimum: (-2.89, -108.85)
Maximum: (2.08, 26.13)
To graph the polynomial y = 2x^3 - 3x^2 - 12x - 32 in the given viewing rectangle [−5, 5] by [−60, 30], we can follow these steps:
1. Plot the x and y-axis on a graph paper.
2. Label the x-axis from -5 to 5 and the y-axis from -60 to 30.
3. Calculate the y-coordinate for different x-values using the polynomial equation y = 2x^3 - 3x^2 - 12x - 32.
4. Plot the points on the graph accordingly.
5. Draw a smooth curve connecting these points.
To find the coordinates of the local extrema (minimum and maximum), we need to find the critical points of the function. A critical point occurs where the derivative of the function equals zero or is undefined.
First, let's find the derivative of the polynomial:
dy/dx = 6x^2 - 6x - 12
Now, let's find the critical points by setting the derivative equal to zero and solving for x:
6x^2 - 6x - 12 = 0
Next, we can solve this quadratic equation:
Using the quadratic formula, x = [-b ± √(b^2 - 4ac)] / 2a
a = 6, b = -6, c = -12
x = [-(-6) ± √((-6)^2 - 4(6)(-12))] / 2(6)
x = [6 ± √(36 + 288)] / 12
x = [6 ± √324] / 12
x = [6 ± 18] / 12
x = (6 + 18) / 12 or x = (6 - 18) / 12
x = 2 or x = -1
Now that we have found the critical points, we can find the corresponding y-coordinates by substituting these x-values into the original polynomial equation:
For x = 2: y = 2(2^3) - 3(2^2) - 12(2) - 32
= 16 - 12 - 24 - 32
= -52
So, the first critical point is (2, -52).
For x = -1: y = 2(-1^3) - 3(-1^2) - 12(-1) - 32
= -2 + 3 + 12 - 32
= -19
So, the second critical point is (-1, -19).
Now, let's determine which of these critical points are local extrema by checking the value of the second derivative (d^2y/dx^2) at these points.
The second derivative is given by:
d^2y/dx^2 = 12x - 6
For x = 2: d^2y/dx^2 = 12(2) - 6
= 18
Since the second derivative is positive, the function has a minimum at (2, -52).
For x = -1: d^2y/dx^2 = 12(-1) - 6
= -18
Since the second derivative is negative, the function has a maximum at (-1, -19).
Therefore, the minimum point is (2, -52) and the maximum point is (-1, -19) for the given polynomial.