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Kipish [7]
3 years ago
7

Explain why the x-coordinates of the points where the graphs of the equations y = 2−x and y = 8x+4 intersect are the solutions o

f the equation 2−x = 8x+4. part
b.make tables to find the solution to 2−x = 8x+4. take the integer values of x between −3 and 3. part
c.how can you solve the equation 2−x = 8x+4 graphically?

Mathematics
1 answer:
Luda [366]3 years ago
4 0
Part 1)
We have two lines:  y = 2-x   and   y = 8x+4
Given two simultaneous equations that are both required to be true.
the solution is the points where the lines cross
Which is where the two equations are equal
Thus the solution that works for both equations is when
2-x = 8x+4
because
where that is true is where the two lines will cross and that is the common point that satisfies both equations.

Part 2) Make tables to find the solution to 2−x = 8x+4. take the integer values of x between −3 and 3

see the attached table
The table shows that none of the integers from [-3,3] work because in no case does
<span>2-x = 8x+4
</span>
To find the solution we need to rearrange the equation to the form x=n
2-x =8x+4
8x+x=2-4
9x=-2
x=-2/9

 The only point that satisfies both equations is
where
 x = -2/9
Find y:
   y = 2-x  = 2 - (-2/9) = 2 + 2/9 = 20/9
Verify we get the same in the other equation
y = 8x+4   =  8(-2/9) + 4 = 20/9 
Thus the only actual solution, being the point where the lines cross,
 is the point (-2/9, 20/9)------> (-0.22,2.22)

Part 3) how can you solve the equation 2−x = 8x+4 graphically?
<span>The point on the graph where the lines cross is the solution to the system of equations
</span>
using a graph tool
see the attached figure

the solution is the point (-0.22,2.22)

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we have been asked to verify that -5, 1/2, and 3/4 are the zeroes of the cubic polynomial 4x^3+20x^2+2x-3

To verify that whether the given values are zeros or not we will substitute the values in the given Polynomial, if it will returns zero, it mean that value is Zero of the polynomial. But if it return any thing other than zeros it mean that value is not the zero of the polynomial.

Let f(x)=4x^3+20x^2+2x-3\\\\\text{when x=-5}\\\\f(-5)=4(-5)^3+20(-5)^2+2(-5)-3=-13\\\\\text{when x=}\frac{1}{2}\\

f( \frac{1}{2} ) = 4 ( \frac{1}{2} )^3+20(\frac{1}{2})^2+2(\frac{1}{2})-3=\frac{7}{2}\\\\

\text{when x=}\frac{3}{4}\\\\

f( \frac{3}{4} ) = 4 ( \frac{3}{4} )^3+20(\frac{3}{4})^2+2(\frac{3}{4})-3=\frac{183}{16}\\

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Since sum of roots=\frac{-b}{a}= \frac{-20}{4}=-5\\

But -5+\frac{1}{2}+\frac{3}{4}=  \frac{-15}{4}\neq-5

Hence we do not find any relation between the coefficients and zeros.

Anyway if the given values doesn't represents the zeros then those given values will not have any relation with the coefficients of the p[polynomial.

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