Given the function image y = KX + (3K-1) + 2, where k is a real number, it is proved that no matter what the value of K is, the image of the original function must pass through these two points RT

Given the function image y = KX + (3K-1) + 2, where k is a real number, it is proved that no matter what the value of K is, the image of the original function must pass through these two points RT

It's y = KX & #178; + (3K-1) x + 2
Let x = 0 get y = 2, the function image must pass through point (0,2)
Let x = - 3
y=k(-3)²+(3k-1)(-3)+2
=9k-9k+3+2
=5, the function image must pass the point (- 3,5)
In conclusion, no matter what the value of K is, the function image must pass through two fixed points: (0,2), (- 3,5)

Given the square of the equation KX about X - (3K-1) x + 2 (K-10) = 0, we prove that no matter what the real number k is, the equation always has real roots

(3K-1) ^ 2-8k (k-1) = 9K ^ 2 - 6K + 1-8k ^ 2 + 8K = k ^ 2 + 2K + 1 = (K + 1) ^ 2 > = 0 (K-10) should be (k-1)