Question

Determine as raizes e as coordenadas do vertice da funçao f(x)=x(ao quadrado) + x - 30

Answer 1

$f(x)= x^{2} +x-30 \\ f(x)=0 \\ \\ x^{2} +x-30=0 \\ (x+ \frac{1}{2}) ^{2} =30+ \frac{1}{4} \\ \\ (x+ \frac{1}{2})^2= \frac{120+1}{4} \\ \\ \sqrt{(x+ \frac{1}{2})^2 } =+ - \sqrt{\frac{121}{4} } \\ \\ x'= \frac{-1+11}{2} \\ \\ x'=5 \\ \\ x''= \frac{-1-11}{2} \\ \\ x''=-6$

Vértice

$f'(x)= \lim_{h \to \ 0} \frac{ f(x+h)-f(x)}{h} \\ f'(x)=\lim_{h \to \ 0 } \frac{(x+h)^2+(x+h)-30- x^{2} -x+30}{h \\ } \\ \\ f'(x)= \lim_{h \to \ 0 } \frac{ 2xh+ h^{2}+x+h }{h} \\ f'(x)= \lim_{h \to \ 0 } \frac{h(2x+h+1)}{h} \\ \\ f'(x)=2x+1$

$f'(x)=0= Xv \\ 2x+1=0 \\ x= \frac{-1}{2} \\ \\ Xv= \frac{-1}{2}$

substituir o valor do Xv para descobrir o Yv
$( \frac{-1}{2})^2+ \frac{-1}{2}-30= Yv \\ \\ \frac{1-2-120}{4}= Yv \\ Yv= \frac{-121}{4}$
Xv= - 1/2                 x'=5
Yv= -121/4             x''=-6