Question

esboce o gráfico da função abaixo:

$f(x) = 4x - 3$
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Answer 1

$\Large\boxed{\begin{array}{l}\underline{\rm func_{\!\!,}\tilde ao\,de\,1^o\,grau}\\\sf \acute E\,toda\,func_{\!\!,}\tilde ao\,que\,assume\,a\,forma\\\sf f(x)=ax+b,com~a\ne0.\\\sf o\,termo~~a~~\acute e\,chamado\,de\,coeficiente\,angular\\\sf e\,o\,termo\,b\,de\,coeficiente\,linear.\\\sf quando\,a > 0\longrightarrow func_{\!\!,}\tilde ao\,crescente\\\sf quando\,a < 0\longrightarrow func_{\!\!,}\tilde ao\,decrescente\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm Construc_{\!\!,}\tilde ao\,do\,gr\acute a fico:}\\\sf Para\,construir\,o\,gr\acute afico\,basta\\\sf determinar\,dois\,pontos:\\\sf 1^o\,intersecc_{\!\!,}\tilde ao\,da\,reta\,com\,o\,eixo\,x\\\boldsymbol{ fac_{\!\!,}a\,y=0\,e\,determine\,x}\\\sf 2^o\,intersecc_{\!\!,}\tilde ao\,da\,reata\,com\,o\,eixo\,y\\\boldsymbol{fac_{\!\!,}a\,x=0\,e\,determine\,y}\end{array}}$

$\Large\boxed{\begin{array}{l}\sf f(x)=4x-3\\\underline{\rm Intersecc_{\!\!,}\tilde ao\,com\,o\,eixo\,x:}\\\sf f(x)=0\\\sf 4x-3=0\\\sf 4x=3\\\sf x=\dfrac{3}{4}\implies A\bigg(\dfrac{3}{4},0\bigg)\\\underline{\rm intersecc_{\!\!,}\tilde ao\,com\,o\,eixo\,y}\\\sf f(0)=4\cdot0-3\\\sf f(0)=-3\implies B(0,-3)\\\sf note\,que\,a=4 > 0\implies func_{\!\!,}\tilde ao\,crescente\end{array}}$