a. y=x^2- 4x+3
b. y=x^2+4x
2007-03-18 23:38:58 · 4 answers · asked by Silver L 1 in Science & Mathematics ➔ Mathematics
a. y=x^2- 4x+3
= x^2 - 4x + (4/2)^2 - (4/2)^2 +3
= x^2 - 4x + 4 - 4 + 3
= (x-2)^2 -1
turning point = (2,-1) which is also the minimum point
b. y=x^2+4x
= x^2 + 4x + 4 - 4
= (x-2)^2 - 4
turning point = (2,-4) which also the minimum point
2007-03-18 23:43:50 · answer #1 · answered by the DoEr 3 · 0⤊ 0⤋
Question a)
y = (x² - 4x + 4) - 4 + 3
y = (x - 2)² - 1
Turning point is (2, - 1)
Question b)
y = (x² + 4x + 4) - 4
y = (x + 2)² - 4
Turning point is (- 2,- 4)
2007-03-19 06:53:38 · answer #2 · answered by Como 7 · 0⤊ 0⤋
a. y=(x-3)(x-1)
b. y=x(x+4)
2007-03-19 06:42:29 · answer #3 · answered by blighmaster 3 · 0⤊ 1⤋
turning point form
y= (x-2)^2 -1
turning point is (-2,-1)
y=(x-2)^2 -4
(-2,-4) is the turning point
were revising this too
2007-03-19 06:49:47 · answer #4 · answered by mayeham7 1 · 0⤊ 0⤋