Asap ! which equation could be the equation of the graphed parabola? a y = -x² - 4x - 5 b y = x² - 4x + 5 c y = x² - 4x - 5

C

y = x² - 4x - 5

Step-by-step explanation:

looking at the graph it crosses the x axis at 5 and -1

using the zero product property in reverse we can find the equation

subtract the roots from x and set equal to zero

(x-a) (x-b) =0  where a and b are the roots of the quadratic

(x-5) (x--1) = 0

(x-5) (x+1) =0

Foil

first x*x=x^2

outer : x*1=1x

inner: -5*x=-5x

last : -5*1=-5

add these together

x^2 +x-5x-5=0

x^2 -4x-5 =0

now set this equal to y to find points outside the x axis

y = x^2 -4x-5

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)