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3 years ago

Help please!!!!!!!!!

Mathematics

1 answer:

dmitriy555 [2]3 years ago

5 0

Answer:

$f\left(0\right)=\left(0\right)^2-4\left(0\right)=0-0=0$

$f\left(1\right)=\left(1\right)^2-4\left(1\right)=1-4=-3$

$f\left(2\right)=\left(2\right)^2-4\left(2\right)=4-8=-4$

$f\left(3\right)=\left(3\right)^2-4\left(3\right)=9-12=-3$

$f\left(4\right)=\left(4\right)^2-4\left(4\right)=16-16=0$

Step-by-step explanation:

Given the function

$f\left(x\right)=x^2-4x$

Putting x=0 in the equation to find f(0)

$f\left(x\right)=x^2-4x$

$f\left(0\right)=\left(0\right)^2-4\left(0\right)=0-0=0$

Putting x=1 in the equation to find f(1)

$f\left(x\right)=x^2-4x$

$f\left(1\right)=\left(1\right)^2-4\left(1\right)=1-4=-3$

Putting x=2 in the equation to find f(2)

$f\left(x\right)=x^2-4x$

$f\left(2\right)=\left(2\right)^2-4\left(2\right)=4-8=-4$

Putting x=3 in the equation to find f(3)

$f\left(x\right)=x^2-4x$

$f\left(3\right)=\left(3\right)^2-4\left(3\right)=9-12=-3$

Putting x=4 in the equation to find f(4)

$f\left(x\right)=x^2-4x$

$f\left(4\right)=\left(4\right)^2-4\left(4\right)=16-16=0$

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