SOLUTION: find all real zeros of the function algebraically ie {{{f(x)=x^5+x^3-6x}}}. analyze the graph of the function algebraically ie {{{g(x)=-x^2+10x-16}}}.

Algebra ->  Functions -> SOLUTION: find all real zeros of the function algebraically ie {{{f(x)=x^5+x^3-6x}}}. analyze the graph of the function algebraically ie {{{g(x)=-x^2+10x-16}}}.       Log On


   

Question 185101This question is from textbook college algebra concepts and models
: find all real zeros of the function algebraically ie f%28x%29=x%5E5%2Bx%5E3-6x.
analyze the graph of the function algebraically ie g%28x%29=-x%5E2%2B10x-16. This question is from textbook college algebra concepts and models

Answer by ZeratuelX(5) About Me  (Show Source):
You can put this solution on YOUR website!
Finding all real zeros means that we are finding all possible values for x that will satisfy the equation and still are in the "real number system".
If we have the function f%28x%29=x%5E5%2Bx%5E3-6x we are looking for all of the x-intercepts, which are the same as roots/zeros. The best way to find these algebraically is to factor it out. In this function there is an x we can factor out from the beginning.
y+=+x%2A%28x%5E4%2Bx%5E2-6%29
Now we are looking for 2 numbers that add up to be 1, and multiply to be -6. The two numbers I came up with are -2 and 3. Let's factor this out using the reverse FOIL method and these numbers.
y+=+x%2A%28x%5E2%2B3%29%2A%28x%5E2-2%29
Since we are finding the zeros, which are essentially roots/x-intercepts, we can set y = 0 because the x-intercept will have a y coordinate of 0. We now have this equation:
0+=+x%2A%28x%5E2%2B3%29%2A%28x%5E2-2%29
Now by the zero factor theorem, if the product is 0, than we can set each factor to 0.
x+=+0
x%5E2%2B3+=+0
x%5E2-2+=+0
Now we solve for x in each of these.
x+=+0 Solved.
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x%5E2%2B3+=+0
x%5E2+=+-3
x+=+sqrt%28-3%29 2 Answers.
x+=+-sqrt%28-3%29 Solved.
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x%5E2-2+=+0
x%5E2+=+2
x+=+sqrt%282%29 2 Answers.
x+=+-sqrt%282%29 Solved.
The first equation is already solved, we already have 1 real root. The second equation can't be done in the real number system since we cannot take the square root of a negative number without using the imaginary number system. The third equation has two possible values, both of which are real and satisfy the equation.
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In the original equation we had x%5E5. Since it is raised to the 5th power, there are 5 total solutions, and we have found 3 of them to be real zeroes. The other two are imaginary numbers (from case 2). So our answer would be:
x+=+0
x+=+sqrt%282%29 Note that these numbers are irrational, but still real.
x+=+-sqrt%282%29 This is because they represent a decimal, that is between two numbers.
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For your second question, we need to graph it and describe the graph. So the equation would look like this:
g%28x%29=-x%5E2%2B10x-16
+graph%28+330%2C+300%2C+-1%2C+10%2C+-10%2C+10%2C+-x%5E2%2B10x-16%29+
We can see that the graph opens down. Since it is a parabola, it starts going up, but it hits a vertex at (5,9) and then starts to go down.