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# hlp plz asap

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i really really need help plz can anybbody help due tommmorow

Feb 22, 2022

#1
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(a)

Because the curve C in Figure 3 crosses the x-axis at points A and B, the solutions of the equation are A and B:

y = x(x + 4)(x - 2)

0 = x(x + 4)(x - 2) ...

now solve for x -> x1 = 0, x2 = -4 and x3 = 2

We already know A or B are not zero,

so A = -4 (because A is negative) and B = 2 (because B is positive).

Feb 23, 2022
edited by Straight  Feb 23, 2022
edited by Straight  Feb 23, 2022
#2
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(b)

y = x(x + 4)(x - 2)    | distribute

y = x(x2 - 2x + 4x - 8)

y = x(x2 + 2x - 8)      | distribute

y = x3 + 2x2 - 8x

now use integration

$$\int (x^3+2x^2-8x)dx$$

now do it yourself.

Feb 23, 2022