Please do questions 18-20
1 answer:
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The intercepts of the third degree polynomial corresponds to the zeros of the equation
y = d*(x-a)*(x-b)(x-c)
Where a, b and c are the roots of the polynomial and d an adjustment coefficient.
y = d*(x+2)*(x)*(x-3)
Lets assume d = 1, and we get
y = (x+2)*(x)*(x-3) = x^3 - x^2 - 6x
We graph the equation in the attached file.
y = d*(x-a)*(x-b)(x-c)
Where a, b and c are the roots of the polynomial and d an adjustment coefficient.
y = d*(x+2)*(x)*(x-3)
Lets assume d = 1, and we get
y = (x+2)*(x)*(x-3) = x^3 - x^2 - 6x
We graph the equation in the attached file.
Answer:
The answer for (a) is at the picture
(b) The vertex is (-2,7)
(c) Yes the parabola open downward
(d) -(x+2)^2+7 the turning point is at (-2,7) since it is where the group curves
Step-by-step explanation:
