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Deffense [45]
3 years ago
6

Plz help ASAP please please

Mathematics
1 answer:
jeyben [28]3 years ago
5 0
It should be figure B

Hope it helps :)
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3 years ago
Which equation could generate the curve in the graph below?
dimulka [17.4K]

Answer:

The answer is option 1.

Step-by-step explanation:

In order to find the equation, you have to apply Discriminant Law, D = b² - 4ac. When D < 0, the equation has no real roots (no x-intercept). When D = 0, it has 2 and equal roots (1 x-intercept). When D > 0, it has 2 distinct roots (2 x-intercept).

Option 1,

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

D =  {( - 2)}^{2}  - 4(3)(1)

D =  - 8

D < 0

Option 2,

y = 3 {x}^{2}  - 6x + 3

D =  {( - 6)}^{2}  - 4(3)(3)

D = 0

Option 3,

y = 3 {x}^{2}  - 7x + 1

D =  {( - 7)}^{2}  - 4(3)(1)

D = 37

D > 0

Option 4,

y = 3 {x}^{2}  - 4x - 2

D =  {( - 4)}^{2}  - 4(3)( - 2)

D = 40

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As you look at the graph, the curve does not meet x-axis so the discriminant must be less than 0.

7 0
3 years ago
Use distributive property to factor method 1. Use distributive property to expand method 2.
sasho [114]

Step-by-step explanation:

I guess method 1 means to deal with whole factors.

x + 5 = (x - 2)(x + 5)

for (x + 5) <> 0 we can divide both sides by this factor :

1 = x - 2

x = 3

for the second solution we deal with

x + 5 = 0

x = -5

so, for x = -5 and x = 3 both functions deliver the same output, and these are the intersection points.

method 2 : we multiply the expression out and solve it then

x + 5 = (x - 2)(x + 5)

x + 5 = x² + 5x - 2x - 10 = x² + 3x - 10

0 = x² + 2x - 15

the general solution to such a square equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 1

b = 2

c = -15

x = (-2 ± sqrt(2² - 4×1×-15))/(2×1) =

= (-2 ± sqrt(4 + 60))/2 = (-2 ± sqrt(64))/2 = (-2 ± 8)/2 =

= -1 ± 4

x1 = -1 + 4 = 3

x2 = -1 - 4 = -5

and you get the 2 solutions again. as expected, they are the same as with method 1, of course.

4 0
2 years ago
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