Jackson Wan's answer to lucas's Secondary 4 A Maths Singapore question.
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y=2x^2+6x-1
use completing the square
y=2(x+3/2)^2-2(3/2)^2-1
=2(x+3/2)^2-9/2-1
=2(x+3/2)^2-11/2.
when x=-3/2, y=-11/2
Vertex= ( -3/2, -11/2)
axis od symmetry x= -3/2
y-intercept(when x=0) = 2(9/4)-11/2 = -1
new curve:f'(x)=
f(x)*2*-1 +3= -4x^2-12x+1
new vertex: (-3/2, 13)
hope this helps
use completing the square
y=2(x+3/2)^2-2(3/2)^2-1
=2(x+3/2)^2-9/2-1
=2(x+3/2)^2-11/2.
when x=-3/2, y=-11/2
Vertex= ( -3/2, -11/2)
axis od symmetry x= -3/2
y-intercept(when x=0) = 2(9/4)-11/2 = -1
new curve:f'(x)=
f(x)*2*-1 +3= -4x^2-12x+1
new vertex: (-3/2, 13)
hope this helps
Date Posted:
2 years ago
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