Question

1 ) Calcule a e b de modo que o gráfico da função definida por y=ax² +bx-3 tenha o vértice no ponto (1,4)

Answer 1

Parábola : $\text a.\text x^2 + \text b\text x +\text c$

vértice :

$\displaystyle \text x_\text v = \ \frac{-\text b}{2.\text a} \\ \\ \displaystyle \text y_\text v = \frac{-(\text b^2-4.\text{a.c})}{4.\text a}$

Temos a parábola y = ax² +bx-3. queremos a e b de modo que o vértice seja (1,4), ou seja :

$\text x_\text v \to \displaystyle \frac{-\text b}{2.\text a} = 1 \to \text b = -2\text a$

$\displaystyle \text y_\text v \to \frac{-(\text b^2-4.\text{a}(-3))}{4.\text a} = 4 \to \text b^2 + 12\text a = -16.\text a$

substituindo b = -2a :

$(2\text a)^2 + 12\text a = -16.\text a$

$4\text a^2 + 12\text a +16.\text a=0$

$4\text a^2 + 28.\text a=0$

$\text a(4\text a + 28 )=0$

$\text a = 0$ ( não convém )

$\displaystyle 4\text a + 28 = 0 \to \text a = \frac{-28}{4}$

$\huge\boxed{\text a = -7} \ \checkmark$

Achando o b :

$\text b = -2.\text a \to b = -2(-7)$

$\huge\boxed{\text b = 14} \ \checkmark$