Find the vertex of the parabola y=2x^2+8x-9

The vertex is (-2,-17).

Step-by-step explanation:

We have given y=2x²+8x-9

To find the x-coordinates we use:

xv= -b/2a

where a=2 and b=8

Now put the values in the formula:

xv= -(8)/2(2)

xv=-8/4

xv= -2

Now to find the y coordinates, we will simply substitute the value in the given equation:

y=2x²+8x-9

y=2(-2)²+8(-2)-9

y=2(4)-16-9

y=8-16-9

y=-8-9

y= -17

Therefore the vertex is(-2, -17)

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

b

step-by-step explanation: