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satela [25.4K]
3 years ago
12

If f(x) = x2 − 1, and f(2a) = 35, then what could be the value of a ?

Mathematics
1 answer:
SVEN [57.7K]3 years ago
4 0
F(2a)=35
35=(2a)^2-1
35=4a^2-1
minus 35 from both sidse
0=4a^2-36
facor out 4
0=4(a^2-9)
factor differece of 2 perfect squares
0=4(a-3)(a+3)
set to zero
a-3=0
a=3

a+3=0
a=-3


a=-3 or 3
hope that's what you're looking for.
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Give an example of a polynomial that has the asymptotes of x=5 and y=0.5
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the horizontal asymptote when the numerator and the denominator have the same degree (in this case, both of a degree of 2) is ration of the coefficients of the numerator and denominator. In this case, the coefficient for numerator x² is 1, and the coefficient for the denominator 2x² is 2, so the horizontal asymptote is y=1/2=0.5

the vertical asymptote is the x value. the denominator cannot be zero, if x²-10x+25=0, x would be 5, so the vertical asymptote is x=5 

this is just one example. There can be others:
(2x²+5x+2)/[(4x-7)(x-5)]  for another example, but this example has a second vertical asymptote 4x-7=0 =>x=7/4
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(9x+2)(4x^2+35x-9)=0
pychu [463]
(9x+2)*(4x^2+35x-9)=0

we can make this...

9x+2=0

and

4x^2+35x-9=0

then...

9x+2=0

\boxed{x=-\frac{2}{9}}


4x^2+35x-9=0

x=\frac{-b\pm\sqrt{b^2-4*a*c}}{2*a}

x=\frac{-35\pm\sqrt{35^2-4*4*(-9)}}{2*4}

x=\frac{-35\pm\sqrt{1369}}{8}

x=\frac{-35\pm37}{8}

x_1=\frac{-35+37}{8}=\frac{1}{4}

x_2=\frac{-35-37}{8}=-9

\boxed{\boxed{S=\{\frac{1}{4},-\frac{2}{9},-9\}}}

I'm sure about my answer, don't matter the order...
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