f(x)=-x^2-5x-6
2006-09-13 02:23:30 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
y=-x^2-5x-6
(n, 0) will intersect the x-axis so y should be equal to zero...
0=-x^2-5x-6
x^2+5x+6=0
(x+3)(x+2)=0
There are two conditions:
0(x+2)=0......x+3=0......x=-3
(x+3)0=0......x+2=0......x=-2
so the x-intercepts would be:
(-3, 0) and (-2, 0)
2006-09-13 02:56:57 · answer #1 · answered by Lin 2 · 0⤊ 0⤋
The x-intercept is where your graph crosses the x-axis. In other words, where y=0. In other words, where f(x)=0.
Soo... you need to solve the equation 0 = -x^2 -5x -6
To get rid of annoying minus signs, multiply through by -1.
0 = x^2 +5x +6
This factors very nicely into:
0 = (x+3)*(x+2)
The only way a product can be zero is if one of the things you're multiplying is zero. In other words, either (x+3) or (x+2) must equal zero.
Therefore, x=-3 or x=-2.
Your x-intercepts are x=-3 and x=-2.
If you have trouble factoring, you can solve the quadratic equation using the quadratic formula. However, factoring is much easier. :-)
2006-09-13 02:40:48 · answer #2 · answered by Bramblyspam 7 · 0⤊ 0⤋
Assuming wide-spread conventions, i.e. x-axis is horizontal, y-axis is vertical: This line does not go the y-axis, except at +/- infinity . because the slope is calculated by employing dividing y/x (for a instantly line), the calculation turns into infinity/0. in case you divide something by employing 0 you get a limiteless quantity. hence the answer to both elements of the question is infinity. a nil intercept occurs even as the line crosses the axis in touch, so it won't be able to be 0 for this reason. a nil gradient (slope) might want to be characterisitic of a horizontal line, the position y=n (n= any variety). this is because the accurate of slope is 0 contained in the y/x calculation and 0 divided by employing something continues to be 0.
2016-11-26 21:09:30 · answer #3 · answered by chittenden 4 · 0⤊ 0⤋
The x-intercept can be easily fould by using "abc equation".
x1 = [-b + (b^2 - 4ac) ^ 0.5] / 2a
x2 = [-b - (b^2 - 4ac) ^ 0.5] / 2a
y = f(x) = ax^2 + bx + c (general form of quadratic equation) so that a = -1, b = -5, c = -6
Substituting a, b ,c value into "abc equation" yields the x intercepts of -3 and -2.
2006-09-13 02:33:03 · answer #4 · answered by Antila 2 · 0⤊ 0⤋
f(x) = -x^2 - 5x - 6
-x^2 - 5x - 6 = 0
-(x^2 + 5x + 6) = 0
x^2 + 5x + 6 = 0
(x + 3)(x + 2) = 0
x = -3 or -2
2006-09-13 07:27:52 · answer #5 · answered by Sherman81 6 · 0⤊ 0⤋
try solving 0=-X^2-5X-6
2006-09-13 02:28:15 · answer #6 · answered by Anonymous · 0⤊ 0⤋
Set f(x) equal to zero and solve. For this particular problem, there are no x-intercepts.
2006-09-13 02:30:08 · answer #7 · answered by bruinfan 7 · 0⤊ 0⤋
x= 6 or -1
{-b+/-(b^2-4ac)^1/2} /2a
a=1 b=-5 c=-6
2006-09-13 02:31:52 · answer #8 · answered by Kaze 1 · 0⤊ 0⤋
x= -3
x= -2
2006-09-13 03:20:14 · answer #9 · answered by shahan_najam 2 · 0⤊ 0⤋