Answer:
it must also have the root : - 6i
Step-by-step explanation:
If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.
This is because in order to render a polynomial with Real coefficients, the binomial factor (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:
![(x-(a+bi))*(x-(a-bi))=\\(x-a-bi)*(x-a+bi)=\\([x-a]-bi)*([x-a]+bi)=\\(x-a)^2-(bi)^2=\\(x-a)^2-b^2(-1)=\\(x-a)^2+b^2](https://tex.z-dn.net/?f=%28x-%28a%2Bbi%29%29%2A%28x-%28a-bi%29%29%3D%5C%5C%28x-a-bi%29%2A%28x-a%2Bbi%29%3D%5C%5C%28%5Bx-a%5D-bi%29%2A%28%5Bx-a%5D%2Bbi%29%3D%5C%5C%28x-a%29%5E2-%28bi%29%5E2%3D%5C%5C%28x-a%29%5E2-b%5E2%28-1%29%3D%5C%5C%28x-a%29%5E2%2Bb%5E2)
where the imaginary unit has disappeared, making the expression real.
So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)
Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.
<h3>
<u>Explanation</u></h3>
We have the given slope value and the coordinate point that the graph passes through.
where m = slope and b = y-intercept. Substitute the value of slope in the equation.
We have the given coordinate point as well. After we substitute the slope, we substitute the coordinate point value in the equation.
<u>Solve</u><u> </u><u>the</u><u> </u><u>equation</u><u> </u><u>for</u><u> </u><u>b-term</u>
The value of b is 6. We substitute the value of b in the equation.
We can also use the Point-Slope form to solve the question.
Given the y1 and x1 = the coordinate point value.
Substitute the slope and coordinate point value in the point slope form.
<u>Simplify</u><u>/</u><u>Convert</u><u> </u><u>into</u><u> </u><u>Slope-intercept</u>
<h3>
<u>Answer</u></h3>
<u>
</u>
Find the absolute value vertex. In this case, the vertex for y=|x−5|y=|x-5| is (5,0)(5,0).
(5,0)(5,0)
The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.
(−∞,∞)(-∞,∞)
{x|x∈R}{x|x∈ℝ}
For each xx value, there is one yy value. Select few xx values from the domain. It would be more useful to select the values so that they are around the xx value of the absolute valuevertex.
xy3241506172
It would be: 5/2930 = 0.0017
