How can you find the x intercepts for the equation
-2(x+4)^2+3
Thanks for your help :)
2007-08-10 19:12:55 · 5 answers · asked by checked.cherry 2 in Science & Mathematics ➔ Mathematics
Sorry everyone-just a mistake
y=-2(x+4)^2+3
2007-08-10 19:32:05 · update #1
- 2 (x + 4)² + 3 = 0 for intercepts
2 (x + 4)² - 3 = 0
2(x² + 8x + 16) - 3 = 0
2x² + 16x + 29 = 0
x = [- 16 ±√(16² - (8 x 29) ] / 4
x = [- 16 ±√(24) ] / 4
x = [- 16 ± 2√6 ] / 4
x = - 4 ± (1/2)√6 are x intercepts.
2007-08-10 19:40:53 · answer #1 · answered by Como 7 · 2⤊ 0⤋
I don't think you will get any answers until you post an EQUATION with an equals sign and a quantity to the right of it.
You have only given half of the question. So it is incomplete.
Please re-post. if you mean this equation:
y = -2(x+4)^2+3 then x intersept means y=0 so
0 = -2(x+4)^2+3 -3 = -2(x+4)^2 (x+4)^2=3/2
(x+4) = +- sqrt(3/2) x = sqrt(3/2) -4 x = -sqrt(3/2) -4
2007-08-11 02:27:34 · answer #2 · answered by 037 G 6 · 0⤊ 0⤋
I assume that this is an equation of the form
y = -2(x+4)^2+3
Set the value of y to zero and solve for x.
Y = 0 (always) at the x intercepts.
0 = -2[x² + 8x + 16] + 3
0 = -2x² - 16x - 32 + 3
0 = -2x² - 16x - 29
0 = - [ 2x² + 16x + 29 ]
Use Quadratic Eqn to solve
x = 2.7753 and
x = 5.2247
2007-08-11 02:21:41 · answer #3 · answered by bedbye 6 · 0⤊ 0⤋
to find the x-intercept, u just need to substitute y=0.
It is because at x-axis, y=0.
Try it yaa
2007-08-11 02:20:49 · answer #4 · answered by Doc^chem 2 · 0⤊ 0⤋
that was our topic in analytic geom... and as far as i know,u need to let y=0, then solve for x....that's all i can share.....
2007-08-11 03:20:02 · answer #5 · answered by anime_luver! 2 · 0⤊ 2⤋