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3 years ago

Help with some algebra please?

1 answer:

7 0

Vertex form is f(x) = a(x - b)^2 + c where the vertex is (b, c) and a is some constant.

So for the first part we have f(x) = a(x + 6)^2 - 1

for root -9:- a( -9+6)^2 - 1 = 0
9a - 1 = 0 , a = 1/9

The other 3 are derived in the same way.
so required equation is f(x) = (1/9)(x + 6)^2 - 1

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Step-by-step explanation:

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The value of x<em> </em>in the polynomial fraction 3/((x-4)•(x-7)) + 6/((x-7)•(x-13)) + 15/((x-13)•(x-28)) - 1/(x-28) = -1/20 is <em>x </em>= 24

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The given equation is presented as follows;

$\frac{3}{(x - 4) \cdot (x - 7) } + \frac{6}{(x - 7) \cdot (x - 13) } + +\frac{15}{(x - 13) \cdot (x - 28) } - \frac{1}{(x - 28) } = - \frac{1}{20}$

Factoring the common denominator, we have;

$\frac{3\cdot(x - 13) \cdot(x - 28) + 6 \cdot(x - 4) \cdot(x - 28) + 15 \cdot(x - 4) \cdot(x - 7) - (x - 4) \cdot (x - 7)\cdot(x - 13)}{(x - 4) \cdot (x - 7)\cdot(x - 13) \cdot(x - 28)} + = - \frac{1}{20}$