For the given quadratic equation convert into vertex form, find the vertex, and find the value for x = 6. y = -2x2  + 2x +2 show the steps so i can see how you got your answer and understand it a little more. you! : ) correct answer will get branliest answer

we have that

y = -2x² + 2x +2

Group terms and move the constant to the opposite side of the equation

y -2 = -2x² + 2x

Factor the leading coefficient

y -2 = -2(x²-x )

Complete the square. Remember to balance the equation by adding the same constants to each side

y -2 -0.25*2= -2(x²-x+0.25 )

y -2.50= -2(x²-x+0.25 )

Rewrite as perfect squares

y -2.50= -2(x-0.50)²

Part a) The equation of a parabola in vertex form is. y=a(x−h)2+k

so

y -2.50= -2(x-0.50)²> y = -2(x-0.50)²+2.50

the answer Part a) is

y = -2(x-0.50)²+2.50

part b) Find the vertex

we know that the vertex is the point (h,k)

so

(h,k) is the point> (0.50,2.50)

the answer part b) is

the vertex is the point (0.50,2.50)

Part c) find the value for x = 6

y = -2(x-0.50)²+2.50> y = -2(6-0.50)²+2.50> y = -2(5.50)²+2.50

y=-60.50+2.50> y=-58

the answer Part c) is

-58

1,589.63 in.^3

step-by-step explanation:

i hope this you : d

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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