Find the vertex
H(x)= -3x^2+24x-46
4 answers
What is the method you were taught for this procedure, since there are several ways to do this?
Example: 4x^2+8x-3
Y=4(x^2+8x+1)^2-3-4
Y=4(x+1)^2-7
Vertex:(-1,-7)
Y=4(x^2+8x+1)^2-3-4
Y=4(x+1)^2-7
Vertex:(-1,-7)
-3x^2+24x-46
=-3(x^2-8+16) +2
=-3(x-4)^2+2
x=4,y=2
=-3(x^2-8+16) +2
=-3(x-4)^2+2
x=4,y=2
H(x)= -3x^2+24x-46
= -3(x^2 - 8x + ....) - 46
= -3(x^2 - 8x + 16) - 46 + 48
= -3(x - 4)^2 + 2
vertex is (4, 2)
or
for f(x) = ax^2 + bx + c
the x of the vertex is -b/(2a)
for H(x)= -3x^2+24x-46
x of the vertex is -24/-6 = 4
H(4) = -48 + 96 - 46 = 2
vertex is (4,2)
= -3(x^2 - 8x + ....) - 46
= -3(x^2 - 8x + 16) - 46 + 48
= -3(x - 4)^2 + 2
vertex is (4, 2)
or
for f(x) = ax^2 + bx + c
the x of the vertex is -b/(2a)
for H(x)= -3x^2+24x-46
x of the vertex is -24/-6 = 4
H(4) = -48 + 96 - 46 = 2
vertex is (4,2)