Find the following, given the equation y = x2 + 4x - 21
a. Equation in general form,y = a(x – h)2 + k
b. Coordinate of the vertex
c. Y-intercept
d. X-intercept(s)
2007-12-16 05:52:00 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Part a
y = (x² + 4x + 4) - 4 - 21
y = (x + 2)² - 25
Part b
Vertex (- 2 , - 25)
Part c
y- intercept is y = 4 - 25 = - 21
Part d
x - intercepts :-
(x + 2) = ± √25
(x + 2) = ± 5
x = 3 , x = - 7
2007-12-20 02:59:05 · answer #1 · answered by Como 7 · 1⤊ 0⤋
Greetings,
a) complete the square to get
y = x^2 + 4x + 4 - 4 - 21
y = (x + 2)^2 - 25
y = 1(x - (-2))^2 + (-25), a =1, h = -2, k = -25
b) vertex at (h,k) is (-2,-25)
c) x = 0, y = -21
d) y = 0 ,
(x + 2)^2 - 25 = 0
(x+2)^2 = 25
x + 2 = +/-5
x + 2 = 5 or x + 2 = -5
x = 3 or x = -7
Regards
2007-12-16 06:08:18 · answer #2 · answered by ubiquitous_phi 7 · 0⤊ 0⤋
a) 1(x-(-2))^2 + (-25)
b) (-2,-25)
c) (-21)
d) -7 and 3
2007-12-16 06:11:05 · answer #3 · answered by Anonymous · 0⤊ 0⤋
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2016-10-11 10:02:45 · answer #4 · answered by cardejon 4 · 0⤊ 0⤋
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2007-12-16 06:00:30 · answer #5 · answered by Tanner J 1 · 0⤊ 0⤋