y=-x^4+8x^2-4y'=-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})