a/ \(y=\dfrac{1}{4}x^2\\\begin{array}{|c|c|c|}\hline x&-2&-1&0&1&2\\\hline y&1&\dfrac{1}{4}&0&\dfrac{1}{4}&1\\\hline\end{array}\\→\text{Hàm số đi qua điểm}\,\, (-2;1);(-1;\dfrac{1}{4});(0;0);(1;\dfrac{1}{4});(2;1)\\y=\dfrac{1}{2}x+2\\\begin{array}{|c|c|c|}\hline x&-2&-1&0&1&2\\\hline y&1&\dfrac{3}{2}&2&\dfrac{5}{2}&3\\\hline\end{array}\\→\text{Hàm số đi qua điểm}\,\, (-2;1);(-1;\dfrac{3}{2});(0;2);(1;\dfrac{5}{2});(2;3)\)
b/ \(\text{Pt hoành độ giao điểm}\\\dfrac{1}{4}x^2=\dfrac{1}{2}x+2\\↔\dfrac{1}{4}x^2-\dfrac{1}{2}x-2=0\\↔x^2-2x-8=0\\↔x^2-4x+2x-8=0\\↔x(x-4)+2(x-4)=0\\↔(x+2)(x-4)=0\\↔x+2=0\quad or\quad x-4=0\\↔x=-2\quad or\quad x=4\\→y=1\quad or\quad y=4\\\text{Vậy tọa độ giao điểm}\,\, (P)∩(d)≡\{(-2;1);(4;4)\}\)
