(a) If x = -1 what is the value of f(x) ? f(x) =
What point is on the graph?
(b) Is the point (0, -5) on the graph ?
(c) If f(x) = -6 what is x ? x= , x=
What point(s) is on the graph?
(d) Find x-intercepts. x = , x=
(e) Find y-intercept. y=
(f) What is the domain?
f(x)=x^2-x-6?
2015-05-11 8:03 pm
回答 (2)
2015-05-11 8:06 pm
(a) (-1)^2 - (-1) - 6 = 1 + 1 - 6 = -4, the point is (-1, -4)
(b) No, 0^2 - 0 - 6 is not -5
(c)
x^2 - x - 6 = -6
x^2 - x = -6 + 6
x^2 - x = 0
x(x - 1) = 0
x = 0 or x = 1
Points are (0, -6) and (1, -6)
(d)
x^2 - x - 6 = 0
x^2 - 3x + 2x - 6 = 0
x(x - 3) + 2(x - 3) = 0
(x + 2)(x - 3) = 0
x-intercepts are -2 and 3
(e) 0^2 - 0 - 6 = 0 - 0 - 6 = -6
(f) The domain of any polynomial, unless otherwise indicated is: all real numbers
(b) No, 0^2 - 0 - 6 is not -5
(c)
x^2 - x - 6 = -6
x^2 - x = -6 + 6
x^2 - x = 0
x(x - 1) = 0
x = 0 or x = 1
Points are (0, -6) and (1, -6)
(d)
x^2 - x - 6 = 0
x^2 - 3x + 2x - 6 = 0
x(x - 3) + 2(x - 3) = 0
(x + 2)(x - 3) = 0
x-intercepts are -2 and 3
(e) 0^2 - 0 - 6 = 0 - 0 - 6 = -6
(f) The domain of any polynomial, unless otherwise indicated is: all real numbers
2015-05-11 8:11 pm
A) = -4
B) no if you replace x with 0 then y would be -6
C) x=1 and x = 0 if f(x) = -6
D) x ints x =3 x=-2
E)(0,-6) y int y = -6
F) all real numbers
B) no if you replace x with 0 then y would be -6
C) x=1 and x = 0 if f(x) = -6
D) x ints x =3 x=-2
E)(0,-6) y int y = -6
F) all real numbers
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