Question

dada a função y=x²-5x+7 determine o vértice da parábola

Answer 1

$y=x^2-5x+7\\\\\\x^2-5x+7=0\\\\ a=1~;~b=-5~;~c=7\\\\\\ \Delta=b^2-4ac\to~~\Delta=(-5)^2-4.1.7\to~~\Delta=25-28\to~~ \boxed{\Delta=-3} \\\\\\ x_v=\dfrac{-b}{2a}\to~~ x_v=\dfrac{-(-5)}{2.1}\to~~ \boxed{x_v=\dfrac{5}{2}} \\\\\\ y_v=\dfrac{-\Delta}{4a}\to~~ y_v=\dfrac{-(-3)}{4.1}\to~~ \boxed{y_v=\dfrac{3}{4}} \\\\\\\\\\ \large\boxed{ \boxed{ S= \left\{\dfrac{5}{2}~;~\dfrac{3}{4} \right\} }}$

Answer 2

y = x² - 5x + 7; iguale a 0.

x² - 5x + 7 = 0;     a; 1       b; -5       c; 7

Δ = b²-4ac ⇔ Δ = 25 - 28
Δ = -3 

Vértice x = $\frac{-b}{2a}$ =  $\frac{5}{2}$

Vértice y = $\frac{-Δ}{4a}$ =  $\frac{3}{4}$