|x+11| =−1 plz help show steps 2

Mathematics

1 answer:

dezoksy [38]3 years ago

3 0

Answer:

No solutions. You can’t have a negative absolute value.

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Suppose the graph of a cubic polynomial function has the same zeroes and passes through the coordinate (0, -5). Write the equati

amid [387]

In this we know all three zeros and one point from which the graph pass.
So we will let specific cubic polynomial function of the form
$f(x) = a(x - x_1)(x-x_2)(x-x_3)$

As we know zeros are that point where we will get value of function equal to zero. So it is basically in form $(x_n , 0)$

SO in given question zeros are (2 , 0) , (3, 0) and (5,0)
So we can say $x_1 = 2 , x_2 = 3 , x_3 = 5$

So required equation is
$f(x) = a (x-2)(x-3)(x-5)$
              $= a[(x^2 - 2x - 3x + 6)(x-5)]$
              $= a[(x^2 - 5x+6)(x-5)]$
              $= a(x^3 - 5x^2 + 6x- 5x^2 + 25x - 30)$
              $= a(x^3 - 10x^2+31x-30)$
Now we have one point (0 , -5) from which graph passes.
So we say at x = 0 , f(x) = -5
$-5= a (0-0+0 - 30)$
$-5 = -30a$
$a = \frac{-5}{-30} = \frac{1}{6}$
So required equation of cubic polynomial is
$f(x) = \frac{1}{6}(x^3 -10x^2+31x-30)$

For finding y - intercept we simply plugin x = 0 in given equation.
As we know at x = 0 , value of function is -5.
So y - intercept is -5.

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