SOLUTION: Using your calculator graph the function f(x) = x 2 + 2x – 3 over the interval [-5,5]. Approximate any local maxima and local minima. Round your answers to two decimal places.

Algebra ->  Graphs -> SOLUTION: Using your calculator graph the function f(x) = x 2 + 2x – 3 over the interval [-5,5]. Approximate any local maxima and local minima. Round your answers to two decimal places.       Log On


   

Question 154132: Using your calculator graph the function f(x) = x 2 + 2x – 3 over the interval
[-5,5]. Approximate any local maxima and local minima. Round your answers to two decimal places.
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Using your calculator graph the function f(x) = x 2 + 2x – 3 over the interval
[-5,5]. Approximate any local maxima and local minima. Round your answers to two decimal places
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B2x%2B-3+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%282%29%5E2-4%2A1%2A-3=16.

Discriminant d=16 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+16+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%282%29%2Bsqrt%28+16+%29%29%2F2%5C1+=+1
x%5B2%5D+=+%28-%282%29-sqrt%28+16+%29%29%2F2%5C1+=+-3

Quadratic expression 1x%5E2%2B2x%2B-3 can be factored:
1x%5E2%2B2x%2B-3+=+%28x-1%29%2A%28x--3%29
Again, the answer is: 1, -3. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B2%2Ax%2B-3+%29

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For the minimum: set the 1st derivative = to zero.
2x+2 = 0
x = -1
f(-1) = 1 - 2 - 3 = -4
So the minimum is as (-1,-4)