Question

9. O gráfico da função real f, do 1° grau, está representada na figura 1 a seguir.

Sendo g a função real definida por g(x) = ax² + bx+c, com vértice no ponto (-1,-5) e A= 20, representada na figura 2. A partir dessas informações determine o valor do coeficiente a, bec. Determine o valor de f-¹(g(1)).

Answer 1

Resposta:

$a=\frac{5}{4}\\\\b=\frac{5}{2}\\\\c=-\frac{15}{4}\\\\f^-^1(g(1) )=2$

Explicação passo a passo:

$\frac{x}{2} + \frac{y}{3} =1\\\\3x+2y=6\\\\2y=6-3x\\\\y=3-\frac{3}{2}x \\\\x=3-\frac{3}{2} f^1(x)\\\\2x=6-3f^-^1(x)\\\\3f^-^1(x)=-2x+6\\\\f^-^1(x)=-\frac{2}{3}x+2$

$g(x) =ax^2+bx+c\\\\x_v=-\frac{b}{2a} \\\\-1=-\frac{b}{2a} \\\\b=2a\\\\y_V=-\frac{\Delta}{4a} \\\\-5=\frac{-(b^2-4ac)}{4a} \\\\20a=b^2-4ac\\\\x_v=\frac{-b}{2a}\\\\-1=\frac{-b}{2a}\\\\b=2a\\\\20a=(2a)^2-4ac \\\\20a=4a^2-4ac\\\\como ~a~>~0 ,~podemos~dividir~por~a\\\\20=4a-4c\\\\5=a-c$

$c=a-5\\\\x'=-3~~e~~x"=1\\\\g(x)=ax^2+bx+c\\\\a.1^2+b.1+c=0\\\\a+b+c=0\\\\a+2a+c=0\\\\3a+c=0\\\\3a+a-5=0\\\\4a=5\\\\a=\frac{5}{4}\\\\b=2a\\\\b=2*\frac{5}{4} \\\\b=\frac{5}{2} \\\\c=a-5\\\\c=\frac{5}{4} -5\\\\c=-\frac{15}{4} \\\\g(x)=\frac{5}{4}x^2+\frac{5}{2} x-\frac{15}{4}\\\\g(1)=\frac{5}{4}+\frac{5}{2} -\frac{15}{4} \\\\g(1)=\frac{5}{4}+\frac{10}{4} -\frac{15}{4} \\\\g(1)=0\\\\f^-^1(g(1))=f^-^1(0) =-\frac{2}{3} .0+2=~2$