Let f(x)=-3x^2+kx+24,where k is a constant .If the y-coordinate of the vertex is 27, then k =??
+130516 Let f(x)=-3x^2+kx+24,where k is a constant .If the y-coordinate of the vertex is 27, then k =??
Taking the derivative, we have
f'(x) = -6x + k ....and at the vertex, this = 0 so we have..
-6x + k = 0 which implies that k = 6x at this point .....so we have
27 = -3x^2 + (6x)x + 24 = -3x^2 + 6x^2 + 24 simplify
3 = 3x^2
1 = x^2 → x = ± 1
So when x = ± 1, then k = ±6
When x = 1, k = 6 and we have
27 = -3(1)^2 +6(1) + 24 which is true
And when x = -1, k = -6
27 =-3(-1)^2 -6(-1) + 24 which is also true
+130516 Let f(x)=-3x^2+kx+24,where k is a constant .If the y-coordinate of the vertex is 27, then k =??
Taking the derivative, we have
f'(x) = -6x + k ....and at the vertex, this = 0 so we have..
-6x + k = 0 which implies that k = 6x at this point .....so we have
27 = -3x^2 + (6x)x + 24 = -3x^2 + 6x^2 + 24 simplify
3 = 3x^2
1 = x^2 → x = ± 1
So when x = ± 1, then k = ±6
When x = 1, k = 6 and we have
27 = -3(1)^2 +6(1) + 24 which is true
And when x = -1, k = -6
27 =-3(-1)^2 -6(-1) + 24 which is also true
