Sketch given function
y=-x^2(x+6)

  see attached

Step-by-step explanation:

The leading coefficient of this cubic (odd-degree) function is negative, so the overall shape of the curve will be downward-sloping (\). The zeros of the factors are at x=-6 and x=0 (multiplicity 2), so we know the first crossing of the x-axis will occur at x=-6. The graph will touch, but not cross, the x-axis at x=0 and continue descending from there.

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The local minimum will be located 2/3 of the way from the "touch" point to the other x-intercept, so is at x=-4. The value of the function there is ...

  y = -(-4)²(-4+6) = -32

The graph is symmetrical about the inflection point halfway between the local extremes, which point is (-2, -16). Then another point on the curve will be 6 units to the right of the local minimum, at (2, -32).

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?

idkit the do u know da way and ligma consumed ninja

step-by-step explanation: