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Lisa [10]
3 years ago
6

I only need one solved I just need to understand how to do it ASAP. Please help me​

Mathematics
1 answer:
KengaRu [80]3 years ago
7 0

Answer:

#19= -11, #20=4, #21=2, #22= 2, #23= 1/3, #24= -3/5, #25= 1/2

Step-by-step explanation:

#19= 1-3x^2    What you do first is that you do 2^2 to get 4, than multiply i by 3 to get 12, than you subtract 12 from 1( 2^2=4, 4*3, 1-12= -11)

#20=2-3x+x^2     what you do is sub in #, 2-3(-1)+(-1)^2, 2+3-1= (2+3)-1, 5-1=4

#21=x-3x^2      What you do is sub #, -1-3(-1)^2=-1+3=2

#22=x+x^2    sub #, 1+1=2

#23= 1/x+1,     sub #, to get 1/2+1 =1/3

#24=x/2-x,    sub #, to get -3/ 2+3= -3/5

#25=1/squarert of x,  sqrt of 4 is 2, 1/2

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Step-by-step explanation:

The polynomial correlation function of a graphing calculator can work with the table of function values to give you the equation. (See attached)

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If you want to solve this by hand, you can write the equations for the unknown coefficients in the polynomial function, then solve those. First of all, notice that you are given the y-intercept, (0, 6), so you know one of the coefficients already. That leaves a system of 3 equations in 3 unknowns that need to be solved.

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<em>Writing the equations</em>

... ax³ +bx² +cx +6 = y . . . . the generic linear equation in a, b, c that you're writing

For x = -1:

... -a +b -c +6 = 8

For x = 1

... a + b + c + 6 = 6

For x = 2

... 8a +4b +2c +6 = 20

In augmented matrix terms, this set of linear equations looks like ...

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<em>Solving the equations</em>

There are a number of ways to solve these. Again, a graphing calculator often has solution functions, such as reducing a matrix to row-echelon form, that will solve this in a keystroke or two.

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Adding these equations gives ...

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3 years ago
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