Do (P) qua A;B;C, thay tọa độ A, B, C vào pt (P) ta được:

\(\left\{{}\begin{matrix}a+b+c=-1\\4a+2b+c=3\\a-b+c=-3\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}a=1\\b=1\\c=-3\end{matrix}\right.\)

\(\Rightarrow\left(P\right):\) \(y=x^2+x-3\)

\(a\ne0\)

a/ \(\left\{{}\begin{matrix}64a+8b+c=0\\-\frac{b}{2a}=6\\\frac{4ac-b^2}{4a}=-12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}64a+8b+c=0\\b=-12a\\4ac-b^2+48a=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}c=32a\\b=-12a\\4a.\left(32a\right)-\left(-12a\right)^2+48a=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=3\\b=-36\\c=96\end{matrix}\right.\)

\(\Rightarrow y=3x^2-36x+96\)

b/ \(\left\{{}\begin{matrix}c=6\\-\frac{b}{2a}=-2\\\frac{4ac-b^2}{4a}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}c=6\\b=4a\\24a-16a^2=16a\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{2}\\b=2\\c=6\end{matrix}\right.\) \(\Rightarrow y=\frac{1}{2}x^2+2x+6\)

\(\left\{{}\begin{matrix}\dfrac{4ac-b^2}{4a}=1\\4a+2b+c=0\\4a-2b+c=-8\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}4ac-b^2=4a\\4a+2b+c=0\\4b=8\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}b=2\\4ac-4=4a\\4a+4+c=0\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}b=2\\ac-1=a\\c=-4a-4\end{matrix}\right.\)

\(\Rightarrow a\left(-4a-4\right)-1=a\)

\(\Rightarrow4a^2+5a+1=0\) \(\Rightarrow\left[{}\begin{matrix}a=-1\Rightarrow c=0\\a=-\dfrac{1}{4}\Rightarrow c=-3\end{matrix}\right.\)

Vậy có 2 pt (P): \(\left[{}\begin{matrix}y=-x^2+2x\\y=-\dfrac{1}{4}x^2+2x-3\end{matrix}\right.\)

Tìm Parabol (P): y=ax²​+bx+c  đi qua điểm A(1;0) và có tung độ đỉnh bằng -1

4A

5. \(\left\{{}\begin{matrix}a+b+2=5\\4a-2b+2=8\end{matrix}\right.\) \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\) \(\Rightarrow y=2x^2+x+2\)

6. \(\left\{{}\begin{matrix}-\frac{b}{2a}=-2\\\frac{4ac-b^2}{4a}=4\\c=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=4a\\24a-16a^2=16a\\c=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{2}\\b=2\\c=6\end{matrix}\right.\) \(\Rightarrow y=\frac{1}{2}x^2+2x+6\)

7. \(\left\{{}\begin{matrix}c=-1\\a+b+c=-1\\a-b+c=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=1\\b=-1\\c=-1\end{matrix}\right.\) \(\Rightarrow y=x^2-x-1\)

8.

a/ \(AM=\sqrt{2}\)

b/ \(AM=\sqrt{10}\)

c/ Không thuộc đồ thị

d/ Không thuộc đồ thị

Đáp án A đúng

Từ đề bài \(\Rightarrow a>0\) và:

\(\left\{{}\begin{matrix}\frac{4ac-b^2}{4a}=-5\\a+b+c=-1\\c=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b^2=36a\\a+b=-5\\c=4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=\frac{b^2}{36}\\\frac{b^2}{36}+b+5=0\\c=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=1\\b=-6\\c=4\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}a=25\\b=-30\\c=4\end{matrix}\right.\)

a: Vì (P) đi qua A(0;1); B(1;2); C(3;-1) nên ta có hệ phương trình:

\(\left\{{}\begin{matrix}a\cdot0^2+b\cdot0+c=1\\a\cdot1^2+b\cdot1+c=2\\a\cdot3^2+b\cdot3+c=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}c=1\\a+b+1=2\\9a+3b+1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=1\\a+b=1\\9a+3b=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}c=1\\9a+9b=9\\9a+3b=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=1\\6b=11\\a+b=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}c=1\\b=\dfrac{11}{6}\\a=1-\dfrac{11}{6}=-\dfrac{5}{6}\end{matrix}\right.\)

b: Vì (P) đi qua M(0;-1); N(1;0) và P(2;3) nên ta có hệ phương trình:

\(\left\{{}\begin{matrix}a\cdot0^2+b\cdot0+c=-1\\a\cdot1^2+b\cdot1+c=0\\a\cdot2^2+b\cdot2+c=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}c=-1\\a+b-1=0\\4a+2b-1=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=-1\\a+b=1\\4a+2b=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}c=-1\\a+b=1\\2a+b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=-1\\-a=-1\\a+b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=-1\\a=1\\b=0\end{matrix}\right.\)

c: Vì (P) đi qua M(1;-2); N(0;4); P(2;1) nên ta có hệ phương trình:

\(\left\{{}\begin{matrix}a\cdot1^2+b\cdot1+c=-2\\a\cdot0^2+b\cdot0+c=4\\a\cdot2^2+b\cdot2+c=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a+b+c=-2\\c=4\\4a+2b+c=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\a+b=-2-c=-6\\4a+2b=1-4=-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}c=4\\4a+4b=-24\\4a+2b=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\2b=-21\\a+b=-6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}c=4\\b=-\dfrac{21}{2}\\a=-6-b=-6+\dfrac{21}{2}=\dfrac{9}{2}\end{matrix}\right.\)

d: Hoành độ đỉnh là 2 nên -b/2a=2

=>b=-4a(1)

Thay x=3 và y=1 vào (P), ta được:

\(a\cdot3^2+b\cdot3+c=1\)

=>\(9a+3b+c=1\left(2\right)\)

Thay x=-1 và y=2 vào (P), ta được:

\(a\cdot\left(-1\right)^2+b\left(-1\right)+c=2\)

=>a-b+c=2(3)

Từ (1),(2),(3), ta có hệ phương trình:

\(\left\{{}\begin{matrix}b=-4a\\9a+3b+c=1\\a-b+c=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-4a\\9a-12a+c=1\\a+4a+c=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}b=-4a\\-3a+c=1\\5a+c=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-4a\\-8a=-1\\5a+c=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a=\dfrac{1}{8}\\b=-4\cdot\dfrac{1}{8}=-\dfrac{1}{2}\\c=2-5a=2-\dfrac{5}{8}=\dfrac{11}{8}\end{matrix}\right.\)