Find the x value or values at which the parabola y = x^2 + 5x + 4
crosses the x-axis. If the parabola does not cross the x-axis, write
"It does not cross the x-axis".
2007-01-18 08:45:45 · 4 answers · asked by haileybaby331 1 in Science & Mathematics ➔ Mathematics
At y=0, the equation is
0 = x^2 + 5x + 4, then
(x+1)(x+4)=0, therefore
It crosses at the x-axis at x=-1 and x=-4.
2007-01-18 08:49:14 · answer #1 · answered by gebobs 6 · 0⤊ 0⤋
It crosses the x-axis at -1 and -4.
2007-01-18 16:50:33 · answer #2 · answered by PBruins1 1 · 0⤊ 0⤋
y = x^2 + 5x + 4 crosses the x-axis means that y=0 so
x^2+5x+4=0
(x+4)(x+1)=0
x+4=0
x=-4
x+1=0
x=-1
x=-1, -4
2007-01-18 17:12:31 · answer #3 · answered by yupchagee 7 · 0⤊ 0⤋
y = x^2 + 5x + 4
=(x+4)(x+1)
the answer is (0, -4) and (0, -1)
2007-01-18 16:49:31 · answer #4 · answered by VanessaM 3 · 0⤊ 0⤋