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Dima020 [189]
3 years ago
12

I need to graph a parabola. I need to find the vertex and a couple coordinates. y=-x^2+2x-4

Mathematics
1 answer:
PolarNik [594]3 years ago
6 0
Hello : 
<span>y=-x²+2x-4
y = -(x²-2x)-4
y = - (x²-2x+1)+1-4
y = - (x-1)² -3
the vertex is : (1 , -3 )</span>
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Factor complete −2x2 + 10x.
mezya [45]

The factors of the quadratic equation is (x + 1) and (x - 6)

Given,

The quadratic equation; -2x² + 10x + 12

We have to find the factors of this equation using quadratic formula;-

Quadratic formula;- \frac{-b(+-)\sqrt{b^{2} -4ac} }{2a}

Here,

a = -2, b = 10 and c = 12

Now,

\frac{-b(+-)\sqrt{b^{2} -4ac} }{2a} = \frac{-10(+-)\sqrt{10^{2} -4 . -2. 12} }{2.-2} = \frac{-10(+-)\sqrt{100 +96} }{-4} = (-10±14) ÷ -4

Solve for,

- 10 + 14 / -4 = -4/4 = -1

That is, (x + 1)

Solve for,

- 10 - 14 / -4 = -24/-4 = 6

That is,  (x - 6)

Therefore, the factors for the given quadratic equation is (x + 1) and (x - 6)

Learn more about quadratic formula here;

brainly.com/question/9300679

#SPJ1

6 0
1 year ago
Multiply. Check picture.
VashaNatasha [74]

The answer is 3x^4-13x^3-x^2-11x+6.

Solution:

Use algebraic identity: a^m\times a^n=a^{m+n}

For example: x^2\times x=x^{2+1}=x^3

Given expression (x^2-5x+2) and (3x^2+2x+3).

To multiply these equations.

(x^2-5x+2)\times(3x^2+2x+3)

             =x^2(3x^2+2x+3)-5x(3x^2+2x+3)+2(3x^2+2x+3)

             =(3x^4+2x^3+3x^2)+(-15x^3-10x^2-15x)+(6x^2+4x+6)

             =3x^4+2x^3+3x^2-15x^3-10x^2-15x+6x^2+4x+6

Combine like terms together.

             =3x^4+(2x^3-15x^3)+(3x^2-10x^2+6x^2)-15x+4x+6

             =3x^4-13x^3-x^2-11x+6

(x^2-5x+2)\times(3x^2+2x+3)=3x^4-13x^3-x^2-11x+6

Hence the answer is 3x^4-13x^3-x^2-11x+6.

3 0
3 years ago
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