The points (1, 7) and (0, 2) fall on a particular line. what is its equation in slope-intercept form?

Slope-intercept form is y = mx + b

So we can manage to do this:

(y - yo) = m.(x - xo)

Where (x, y) and (xo, yo) are points of this line, this way we can discover the slope.

(7 - 2) = m.(1 - 0)

5 = m

m = 5

So,

y = 5x + b

Now we still have to use (y - yo) = m.(x - xo), but this time we will put only 1 point and the slope.

(y - yo) = m.(x - xo)

(y - 2) = 5.(x - 0)

y - 2 = 5x

y = 5x + 2

So, this is the line in slope-intercept form.

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! : )

cretdit to source

step-by-step explanation:

i found the koficent of the linear function and i foun c acordind to the strict formula y=kx+c and then i took 2 as an x and replacet it on the formula i found and then foun the y