Question

Answer 1

a: Bảng giá trị:

x	-4	-2	0	2	4
\(y=\frac12x^2\)	8	2	0	2	8
y=-x+6	10	8	6	4	2

Vẽ đồ thị:

b: Phương trình hoành độ giao điểm là:

\(\frac12x^2=-x+6\)

=>\(x^2=-2x+12\)

=>\(x^2+2x-12=0\)

=>\(x^2+2x+1-13=0\)

=>\(\left(x+1\right)^2=13\)

=>\(\left[\begin{array}{l}x+1=\sqrt{13}\\ x+1=-\sqrt{13}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\sqrt{13}-1\\ x=-\sqrt{13}-1\end{array}\right.\)

Khi \(x=\sqrt{13}-1\) thì \(y=\frac12\left(\sqrt{13}-1\right)^2=\frac12\left(14-2\sqrt{13}\right)=7-\sqrt{13}\)

Khi \(x=-\sqrt{13}-1\) thì \(y=\frac12\left(-\sqrt{13}-1\right)^2=\frac12\left(14+2\sqrt{13}\right)=7+\sqrt{13}\)