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sergij07 [2.7K]
1 year ago
7

Find a cubic polynomial whose zeros are 1/2,1 and -3.

Mathematics
1 answer:
Alexxx [7]1 year ago
5 0

Answer:

2x³+3x²-8x+3

Step-by-step explanation:

So we know x=½,x=1 and x=-3

Then (2x-1)(x-1)(x+3) are the factors

(2x-1)[(x²+3x-x-3)]

(2x-1)[(x²+2x-3)]

2x³+4x²-6x-x²-2x+3

2x³+3x²-8x+3

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12 cm long and 7 wide
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If (-1, y) lies on the graph of y= 2^2x, then y=<br> -4<br> 1/4<br> 1
AfilCa [17]

Answer:

1/4

Step-by-step explanation:

Substitute x=-1 to find y

Given y= 2^2x

y(-1)=2^-2=1/(2^2)

7 0
2 years ago
Please help me with this please and thank you
Alina [70]

Step-by-step explanation:

y = ax^n + bx^(n-1) ... + z

(0, 1) tells us that the constant term z = 1.

because x = 0 eliminates all other terms. and as the sum is 1, that means z = 1.

for the other pairs I strongly suspect

(-1, 3) : 2×(-1)² + 1 = 2×1 + 1 = 3

(-2, 9) : 2×(-2)² + 1 = 2×4 + 1 = 9

so, the fitting equation is

y = 2x² + 1

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Tcecarenko [31]
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6 0
2 years ago
Write 3x^2-18x-6 in vertex form
Vedmedyk [2.9K]
The standard form of a quadratic equation is \displaystyle{ y=ax^2+bx+c, while the vertex form is:

                      y=a(x-h)^2+k, where (h, k) is the vertex of the parabola.

What we want is to write \displaystyle{ y=3x^2-18x-6 as y=a(x-h)^2+k

First, we note that all the three terms have a factor of 3, so we factorize it and write:

\displaystyle{ y=3(x^2-6x-2).


Second, we notice that x^2-6x are the terms produced by (x-3)^2=x^2-6x+9, without the 9. So we can write:

x^2-6x=(x-3)^2-9, and substituting in \displaystyle{ y=3(x^2-6x-2) we have:

\displaystyle{ y=3(x^2-6x-2)=3[(x-3)^2-9-2]=3[(x-3)^2-11].

Finally, distributing 3 over the two terms in the brackets we have:

y=3[x-3]^2-33.


Answer: y=3(x-3)^2-33
6 0
3 years ago
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