$y=-x^4+8x^2-4$$y'=-4x^3+16x$
$y''=-12x^2+16$
$y''(x_0)=13$ => $-12x^2_0+16=13$ => $x^2_0=\frac{1}{4}$
=> $x_0=\frac{1}{2}$ hoặc $x_0=\frac{-1}{2}$
gọi $M(x_0;y_0)$ là tiếp điểm của (C)
+) $x_0=\frac{1}{2}$ => $y_0=\frac{-33}{16}$
$k=y_0'=\frac{15}{2}$
=> pttt: $y+\frac{33}{16}=\frac{15}{2}(x-\frac{1}{2})$
+) $x_0=\frac{-1}{2}$ => $y_0=\frac{-33}{16}$
$k=y_o'=\frac{-15}{2}$
=> pttt:$y+\frac{33}{16}=\frac{-15}{2}(x+\frac{15}{2})$