Question

Determine os zeros da função f(x)=2x²-3x+1​

Answer 1

Resposta:

(1, 0); (1/2, 0); (0, 1)

Explicação passo-a-passo:

f(x)=2x²-3x+1

para f (x) = 0

2x² - 3x + 1 = 0

x = [3 ± √(3² - 4.2.1)]/2.2

x = [3 ± √(9 - 8)]/4

x = (3 ± 1)/4

x' = (3+1)/4 = 1

x" = (3-1)/4 = 2/4 = 1/2

para x = 0

f(x)=2x²-3x+1

f(x) = y

y = 2.0² - 3.0 + 1

y = 0 - 0 + 1

y = 1

Answer 2

$\Large\boxed{\begin{array}{l}\sf Os~zeros~da~func_{\!\!,}\tilde ao~s\tilde ao~os~valores~de~x\\\sf que~tornam~a~mesma~nula.\end{array}}$

$\large\boxed{\begin{array}{l}\sf f(x)=2x^2-3x+1\\\sf f(x)=0\\\sf 2x^2-3x+1=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-3)^2-4\cdot2\cdot1\\\sf\Delta=9-8\\\sf\Delta=1\\\sf x=\dfrac{3\pm\sqrt{1}}{2\cdot2}\\\\\sf x=\dfrac{3\pm1}{4}\begin{cases}\sf x_1=\dfrac{3+1}{4}=\dfrac{4}{4}=1\\\\\sf x_2=\dfrac{3-1}{4}=\dfrac{2\div2}{4\div2}=\dfrac{1}{2}\end{cases}\\\sf portanto~os~pontos~ser\tilde ao~da~forma~(x,0)\\\sf ou~seja\\\sf A~(1,0)~~B\bigg(\dfrac{1}{2},0\bigg)\end{array}}$