O declive da reta BC \u00e9 ${m_{BC}} =\u00a0 – \\frac{6}{4} =\u00a0 – \\frac{3}{2}$ e a ordenada na origem \u00e9 ${b_{BC}} =\u00a0 – 4$. Logo, $BC:y =\u00a0 – \\frac{3}{2}x – 4$.<\/p>\n O declive da reta CD \u00e9 ${m_{CD}} = \\frac{{12}}{3} = 4$ e a ordenada na origem \u00e9 ${b_{CD}} =\u00a0 – 4$. Logo, $CD:y = 4x – 4$.<\/p>\n O declive da reta DE \u00e9 ${m_{DE}} =\u00a0 – \\frac{8}{4} =\u00a0 – 2$, pelo que a sua equa\u00e7\u00e3o reduzida \u00e9 da forma $y = \\frac{5}{4}x + b$. Dado que o ponto E pertence a esta reta, vem: $0 =\u00a0 – 2 \\times 7 + b \\Leftrightarrow b = 14$. Logo, $DE:y =\u00a0 – 2x + 14$.<\/p>\n Assim, a fun\u00e7\u00e3o $f$ pode ser definida por: \u00a0Por outro lado, vem:![\"Gr\u00e1fico](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/02\/11-2-pag137-5.png\")
<\/a>Considere a fun\u00e7\u00e3o $f$, cuja representa\u00e7\u00e3o gr\u00e1fica se apresenta na figura ao lado.<\/p>\n\n
\n
<\/a>O declive da reta AB \u00e9 ${m_{AB}} = \\frac{5}{4}$, pelo que a sua equa\u00e7\u00e3o reduzida \u00e9 da forma $y = \\frac{5}{4}x + b$. Dado que o ponto B pertence a esta reta, vem: $2 = \\frac{5}{4} \\times \\left( { – 4} \\right) + b \\Leftrightarrow b = 7$. Logo, $AB:y = \\frac{5}{4}x + 7$.\n
\n\\[f\\left( x \\right) = \\left\\{ {\\begin{array}{*{20}{c}}
\n{\\frac{5}{4}x + 7}& \\Leftarrow &{x \\leqslant\u00a0 – 4} \\\\
\n{ – \\frac{3}{2}x – 4}& \\Leftarrow &{ – 4 < x \\leqslant 0} \\\\
\n{4x – 4}& \\Leftarrow &{0 < x \\leqslant 3} \\\\
\n{ – 2x + 14}& \\Leftarrow &{x > 3}
\n\\end{array}} \\right.\\]
\n\u00ad<\/p>\n<\/li>\n
\n\\[\\begin{array}{*{20}{l}}
\n{g\\left( x \\right) = f\\left( {x + 2} \\right) + 1}& = &{\\left\\{ {\\begin{array}{*{20}{c}}
\n{\\frac{5}{4}\\left( {x + 2} \\right) + 7 + 1}& \\Leftarrow &{x + 2 \\leqslant\u00a0 – 4} \\\\
\n{ – \\frac{3}{2}\\left( {x + 2} \\right) – 4 + 1}& \\Leftarrow &{ – 4 < x + 2 \\leqslant 0} \\\\
\n{4\\left( {x + 2} \\right) – 4 + 1}& \\Leftarrow &{0 < x + 2 \\leqslant 3} \\\\
\n{ – 2\\left( {x + 2} \\right) + 14 + 1}& \\Leftarrow &{x + 2 > 3}
\n\\end{array}} \\right.} \\\\
\n{}& = &{\\left\\{ {\\begin{array}{*{20}{c}}
\n{\\frac{5}{4}x + \\frac{{21}}{2}}& \\Leftarrow &{x \\leqslant\u00a0 – 6} \\\\
\n{ – \\frac{3}{2}x – 6}& \\Leftarrow &{ – 6 < x \\leqslant\u00a0 – 2} \\\\
\n{4x + 5}& \\Leftarrow &{ – 2 < x \\leqslant 1} \\\\
\n{ – 2x + 11}& \\Leftarrow &{x > 1}
\n\\end{array}} \\right.}
\n\\end{array}\\]
\n<\/a>O gr\u00e1fico de $g$ pode obter-se a partir do gr\u00e1fico de $f$ por transla\u00e7\u00e3o associada ao vetor $\\overrightarrow u\u00a0 = \\left( { – 2,1} \\right)$.
\n\u00ad<\/li>\n
\n\u00ad<\/li>\n
\n\\[h\\left( x \\right) = \\left\\{ {\\begin{array}{*{20}{c}}
\n{ – \\frac{5}{4}x – 6}& \\Leftarrow &{x \\leqslant\u00a0 – 4} \\\\
\n{\\frac{3}{2}x + 5}& \\Leftarrow &{ – 4 < x \\leqslant 0} \\\\
\n{ – 4x + 5}& \\Leftarrow &{0 < x \\leqslant 3} \\\\
\n{2x – 13}& \\Leftarrow &{x > 3}
\n\\end{array}} \\right.\\]<\/p>\n
\n\\[\\begin{array}{*{20}{l}}
\n{h\\left( x \\right) =\u00a0 – f\\left( x \\right) + 1}& = &{\\left\\{ {\\begin{array}{*{20}{c}}
\n{ – \\left( {\\frac{5}{4}\\left( {x – 0} \\right) + 7} \\right) + 1}& \\Leftarrow &{x – 0 \\leqslant\u00a0 – 4} \\\\
\n{ – \\left( { – \\frac{3}{2}\\left( {x – 0} \\right) – 4} \\right) + 1}& \\Leftarrow &{ – 4 < x – 0 \\leqslant 0} \\\\
\n{ – \\left( {4\\left( {x – 0} \\right) – 4} \\right) + 1}& \\Leftarrow &{0 < x – 0 \\leqslant 3} \\\\
\n{ – \\left( { – 2\\left( {x – 0} \\right) + 14} \\right) + 1}& \\Leftarrow &{x – 0 > 3}
\n\\end{array}} \\right.} \\\\
\n{}& = &{\\left\\{ {\\begin{array}{*{20}{c}}
\n{ – \\frac{5}{4}x – 6}& \\Leftarrow &{x \\leqslant\u00a0 – 4} \\\\
\n{\\frac{3}{2}x + 5}& \\Leftarrow &{ – 4 < x \\leqslant 0} \\\\
\n{ – 4x + 5}& \\Leftarrow &{0 < x \\leqslant 3} \\\\
\n{2x – 13}& \\Leftarrow &{x > 3}
\n\\end{array}} \\right.}
\n\\end{array}\\]<\/p>\n<\/li>\n<\/ol>\n