Identify the equation in slope intercept from that has a slope of 2 and goes through the point (4, 6).

y=−2x+2

y=2x+14

y=2x+6

y=2x−2

  y = 2x−2

Step-by-step explanation:

You are given a slope and a point, so you can write the equation using point-slope form:

  y -k = m(x -h) . . . . . . a line with slope m through point (h, k)

Putting in the numbers given, you have ...

  y -6 = 2(x -4)

  y = 2x -8 +6 . . . . eliminate parentheses, add 6

  y = 2x -2 . . . . . . . collect terms.

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You know that the slope is the coefficient of x, so the first choice is eliminated right away. Trying the point values in the other equations shows you that they only work for the last choice, as shown above.

D) y=2x-2

Step-by-step explanation:

y-y1=m(x-x1)

y-6=2(x-4)

y=2x-8+6

y=2x-2

the correct answer is

c. 4

step-by-step explanation:

the given polynomial is

the degree of the polynomial is the highest exponent which is 4.

the -4 is called the leading coefficient.

this polynomial has only one term so it is a monomial.

the correct answer is option c.

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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