Where is the vertex of the graph of
Y=x^2-5x+3
2006-11-13 08:51:05 · 5 answers · asked by angel l 1 in Science & Mathematics ➔ Mathematics
looks like the vertex is (-5, 3)
2006-11-13 08:53:23 · answer #1 · answered by chickadee_ajm 4 · 1⤊ 1⤋
To find the x-coordinate of the vertex of any parabola
ax^2 + bx + c
, you do
-b/2a
In this case, b = -5 and a = 1:
5/2 = 2.5
Now plug 2.5 into the equation:
Y = (2.5)^2 - 5(2.5) + 3
Y = -3.25
*** The vertex of the graph is (2.5, -3.25)
2006-11-13 16:54:04 · answer #2 · answered by عبد الله (ドラゴン) 5 · 0⤊ 0⤋
Completing the square, you can rewrite this as
y = x^2 - 5x + 3 = (x - 2.5)^2 - 6.25 + 3
or
y + 3.25 = (x - 2.5)^2
So the vertex is at (2.5, -3.25).
2006-11-13 16:59:21 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Y=x^2-5x+3
y'=2x-5=0
sx=5
x=5/2
y=(5/2)^2-5(5/2)+3
y=25/4-25/2+3
y=3-25/4
y=(12-25)/4=-13/4
the vertex is at (5/2, -13/4)
2006-11-13 16:59:06 · answer #4 · answered by yupchagee 7 · 0⤊ 0⤋
2.5,-3.25
2006-11-13 17:06:24 · answer #5 · answered by jerzey79 2 · 0⤊ 0⤋