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andrezito [222]
3 years ago
9

Evaluate f(x)=(2x+1)^2 at -2,-1,0,1 and 2

Mathematics
1 answer:
stiks02 [169]3 years ago
5 0

Answer: at -2=9

at -1= 1

at 0=1

at 1= 9

at 2=25

Step-by-step explanation:

F(x)= (2x + 1)²

at -2

F(x) = (2(-2) + 1)²

= (-4+1)²

=(-3)²

=9.

Same goes with all points

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Answer:

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

f\left(x\right)=x^3-6x^2+3x+10

As the highest power of the x-variable is 3 with the leading coefficients of 1.

  • So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

f\left(x\right)=x^3-6x^2+3x+10

0=x^3-6x^2+3x+10              ∵  f(x)=0

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Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0

so

\left(x+1\right)\left(x-2\right)\left(x-5\right)=0    

Using the zero factor principle

if  ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0

x=-1,\:x=2,\:x=5

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x=-1,\:x=2,\:x=5

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