Question

UUUUUUUUU Questão 02. Dada a função f(x) = x2 - 4x + 4, o valor da f(-3) é: (A) 10. (B) 15. (C) 20. (D) 25.
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Answer 1

Resposta: 25

Explicação passo a passo:

f(x) = x² - 4x + 4

f(-3) = (-3)² -4*(-3) + 4 =

            9 + 12 + 4 = 25

alternativa "D"

Answer 2

$\mathrm{Dominio\:de\:}\:x^2-4x+4\::\quad \begin{bmatrix}\mathrm{Solucao:}\:&\:-\infty \:<x<\infty \\ \:\mathrm{Notacao\:intervalo}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}\\\\\mathrm{Imagem\:de\:}x^2-4x+4:\quad \begin{bmatrix}\mathrm{Solucao:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Notacao\:intervalo}&\:[0,\:\infty \:)\end{bmatrix}\\\\\mathrm{Pontos\:de\:interseccao\:com\:o\:eixo\:de}\:x^2-4x+4:\quad \mathrm{X\:intersepta}:\:\left(2,\:0\right),\:\mathrm{Y\:intersepta}:\:\left(0,\:4\right)$

$\mathrm{Vertice\:de}\:x^2-4x+4:\quad \mathrm{Minimo}\space\left(2,\:0\right)$

$\mathrm{O\:vertice\:de\:uma\:parabola\:com\:concavidade\:para\:cima\:ou\:para\:baixo\:na\:forma}\:y=ax^2+bx+c\:\mathrm{e}\:x_v=-\frac{b}{2a}$

$\mathrm{Os\:parametros\:da\:parabola\:sao:}$

$a=1,\:b=-4,\:c=4\\\\x_v=-\frac{b}{2a}\\\\x_v=-\frac{\left(-4\right)}{2\cdot \:1}\\\\\mathrm{Simplificar}\\\\x_v=2\\\\y_v=0\\\\\mathrm{Portanto,\:o\:vertice\:da\:parabola\:e}\\\\\left(2,\:0\right)\\\\\mathrm{Se}\:a<0,\:\mathrm{entao\:o\:vertice\:e\:um\:valor\:maximo}\\\\\mathrm{Si}\:a>0,\:\mathrm{entao\:o\:vertice\:e\:um\:valor\:minimo}\\\\a=1\\\\\mathrm{Mimo}\space\left(2,\:0\right)$