Find the vertex, zero(s), and y-intercept of the graph of y = 2x2 + 8x – 90. a. vertex: (–2,–98); zeros: (–5,0), (9,0) y-intercept: (0,–90) b. vertex: (2,–98); zeros: (–5,0), (9,0) y-intercept: (0,90) c. vertex: (2,98); zeros: (5,0), (–9,0) y-intercept: (0,–90) d. vertex: (–2,–98); zeros: (5,0), (–9,0) y-intercept: (0,–90)

vertex is (-2,-98)

zeros are (5,0) (-9,0)

y-intercept is (0,-90)

Step-by-step explanation:

we are given

Vertex:

we can use vertex formula

we can compare and find a,b and c

a=2 , b=8 and c=-90

so, we can plug it in formula

now, we can find y-value

so, vertex is (-2,-98)

zeros:

we can set y=0

and then we can solve for x

we can factor it

zeros are

(5,0) (-9,0)

y-intercept:

we can plug x=0 and find y

So, y-intercept is (0,-90)

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