Kishi draws a parabola with vertex at (6, 8) and x-intercepts at (2, 0) and (10, 0). Which is the equation of the parabola in standard form? y=−(12)x2+6x−10

y=−(12)x2−6x−10

y=(12)x+6x−10

y=−(12)x−6x+10

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

,jkulgiurutrjyftkugyi.uykujuyxjdkytuf,yukj

step-by-step explanation:

ℓσℓ ιт ∂σит тяσυвℓє мє αт αℓℓ ιм gℓα∂ тσ нєℓρ ρρℓ συт

ωє ιт ¢αи вє ωяιттєи αѕ

step-by-step explanation:

нσρє ι ¢συℓ∂ нєℓρ уσυ συт