Question

1.Determine a lei de formação da função f se o valor da abscissa do vértice for 1, f(0) = 1 e f(-1) = 4.

a)f(x) = x² - 4x + 2

b)f(x) = x² + 4x + 2

c)f(x) = x² - 2x + 1

d)f(x) = x² + 2x + 1

2.Dada uma função quadrática temos que f(0) = 4, f(1) = 6 e f(2) = 12. Determine sua lei de formação.

a)f(x) = x² + 4x

b)f(x) = 4x² + x

c)f(x) = x² + 4

d)f(x) = 2x² + 4​

Answer 1

1)

$\sf{x_V=1}\\\sf{f(0)=1}\\\sf{f(-1)=4}$

$\sf{x_V=1}\\\sf{-\dfrac{b}{2a}=1}\\\sf{b=-2a}\\\sf{f(x)=ax^2+bx+c}\\\sf{f(x)=ax^2-2ax+c}$

$\sf{f(0)=a\cdot0^2-2a\cdot0+c}\\\sf{c=1}$

$\sf{f(-1)=a\cdot(-1)^2-2a\cdot(-1)+1}\\\sf{a+2a+1=4}\\\sf{3a=4-1}\\\sf{3a=3}\\\sf{3a=3}\\\sf{a=\dfrac{3}{3}}\\\sf{a=1}\\\sf{b=-2\cdot1}\\\sf{b=-2}$

$\large\boxed{\begin{array}{l}\sf{A~lei~da~func_{\! \!,}\tilde{a}o~\acute{e}}\end{array}}$

$\large\boxed{\boxed{\boxed{\boxed{\sf{f(x)=x^2-2x+1}}}}}$

$\large\boxed{\boxed{\boxed{\boxed{\sf{\maltese~~alternativa~~c}}}}}$

$\dotfill$

2)

$\sf{f(0)=4\implies c=4}\\\sf{f(1)=a\cdot1^2+b\cdot1+4}\\\sf{a+b+4=6}\\\sf{a+b=6-4}\\\sf{a+b=2}\\\sf{f(2)=a\cdot2^2+b\cdot2+4}\\\sf{4a+2b+4=12}\\\sf{4a+2b=12-4}\\\sf{4a+2b=8\div2}\\\sf{2a+b=4}$

$\begin{cases}\sf{a+b=2}\\\sf{2a+b=4}\end{cases}}$

$+\underline{\begin{cases}\sf{-a-b=-2}\\\sf{2a+b=4}\end{cases}}$

$\sf{a=2}\\\sf{a+b=2}\\\sf{2+b=2}\\\sf{b=2-2}\\\sf{b=0}$

$\large\boxed{\begin{array}{c}\sf{portanto~a~lei~da~func_{\!\!,}\tilde{a}o~\acute{e}}\end{array}}$

$\large\boxed{\boxed{\boxed{\boxed{\sf{f(x)=2x^2+4}}}}}$

$\large\boxed{\boxed{\boxed{\boxed{\sf{\maltese~~alternativa~~d}}}}}$