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Ostrovityanka [42]
4 years ago
9

3x+2y=12 In function notation

Mathematics
1 answer:
Mariulka [41]4 years ago
6 0

Answer:

f(x) = (-3/2)x + 6

Step-by-step explanation:

Let x be the independent variable and y the dependent one.

Solving 3x+2y=12 for y, we first get 2y = -3x + 12. Then

y = (-3/2)x + 6.  Replacing 'y' with 'f(x)', we have:

f(x) = (-3/2)x + 6, which is 3x+2y=12 when y is the dependent variable.

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NikAS [45]

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(2,-3) and (-2,5)

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Let us graph the two equations one by one.

1. f(x)=-2x+1

If we compare this equation with the slope intercept form of a line which is given as

y=mx+c

we see that m = -1 and c =1

Hence the slope of the line is -2 and the y intercept is 1. Hence one point through which it is passing is (0,1) .

Let us find another point by putting x = 1 and solving it for y

y=-2(1)+1

y=-2+1 = -1

Let us find another point by putting x = 2 and solving it for y

y=-2(2)+1

y=-4+1 = -3

Hence the another point will be (2,-3)

Let us find another point by putting x = -2 and solving it for y

y=-2(-2)+1

y=+1 = 5

Hence the another point will be (-2,5)

Now we have two points (0,1) ,(1,-1) ,  (2,-3) and (-2,5) we joint them on line to obtain our line  

2.

g(x)=y=x^2-2x-3

y=x^2-2x+1-1-3

y=(x-1)^2-4

(y+4)=(x-1)^2

It represents the parabola opening upward with vertices (1,-4)

Let us mark few coordinates so that we may graph the parabola.

i) x=0 ; y=y=(0)^2-2(0)-3=0-0-3=-3 ; (0,-3)

ii)x=-1 ; y=(-1)^2-2(-1)-3=1+2-3=0 ; (-1,0)

iii) x=2 ; y=(2)^2-2(2)-3 = 4-4-3 =-3 ;(2,-3)

iii) x=1 ; y=(1)^2-2(1)-3 = 1-2-3 =-4  ;(1,-4)

iii) x=-2 ; y=(-2)^2-2(-2)-3 = 4+4-3 =5  ;(-2,5)

Now we plot them on coordinate axis and line them to form our parabola

When we plot them we see that we have two coordinates (2,-3) and (-2,5) are common , on which our graphs are intersecting. These coordinates are solution to the two graphs.

3 0
4 years ago
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