Đáp án:

Giải thích các bước giải:

Đáp án:

Giải thích các bước giải:

$a)y = -x^2 - 3x + 4 \\\text{ta có BBT}\\\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}\dfrac{-3}{2}&\text{}&\text{$+\infty$}\\\hline\text{$y$}&&\text{}&\text{}\dfrac{-11}{4}&\text{}&\text{}\\&\text{}&\text{$\nearrow$}&\text{}&\text{}\searrow&\text{}\\&\text{$$}&\text{}&\text{}&\text{}&\text{}\\\hline \end{array}\\b)y = x^2 - 4x + 3\\\text{ta có BBT}\\ \begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}2&\text{}&\text{}&\text{$+\infty$}\\\hline \text{$y$}&&\text{}&\text{}&\text{}&\text{}&\text{}\\&\text{}&\text{$\searrow$}&\text{}&\text{}\nearrow&\text{}&\text{}\\&\text{$$}&\text{}&\text{}-1&\text{}&\text{}&\text{}\\\hline \end{array}\\c)y = x^2 - 2x\\\text{ta có BBT}\\\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}1&\text{}&\text{}&\text{$+\infty$}\\\hline \text{$y$}&&\text{}&\text{}&\text{}&\text{}&\text{}\\&\text{}&\text{$\searrow$}&\text{}&\text{}\nearrow&\text{}&\text{}\\&\text{$$}&\text{}&\text{}-1&\text{}&\text{}&\text{}\\\hline \end{array}\\d)y = -x^2 + 3\\\text{ta có BBT}\\\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}0&\text{}&\text{}&\text{$+\infty$}\\\hline \text{$y$}&&\text{}&\text{}3&\text{}&\text{}&\text{}\\&\text{}&\text{$\nearrow$}&\text{}&\text{}\searrow&\text{}&\text{}\\&\text{$$}&\text{}&\text{}&\text{}&\text{}&\text{}\\\hline \end{array}\\e)y = x^2 + 2x + 3\\\text{ta có BBT}\\\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}-1&\text{}&\text{}&\text{$+\infty$}\\\hline \text{$y$}&&\text{}&\text{}&\text{}&\text{}&\text{}\\&\text{}&\text{$\searrow$}&\text{}&\text{}\nearrow&\text{}&\text{}\\&\text{$$}&\text{}&\text{}2&\text{}&\text{}&\text{}\\\hline \end{array}\\ \mathscr{X_{\textit{in hay nhất}}}$