The x-coordinate of the vertex is precisely halfway between the given roots 2 and 4; thus, x = (2+4)/2, or x=3. The y-coordinate is the value of the function at this x=3.
We can actually determine the equation of this quadratic from the zeros:
f(x) = (x-2)(x-4), or f(x) = x^2 - 6x + 8. To find the y-coord. of the vertex,, subst. 3 for x in f(x) = x^2 - 6x + 8: f(3) = 3^2 - 6(3) + 8 = 9 - 18 + 8, or -1. Then we know that the vertex is at (3, -1).
Answer:
-5x + 3
Step-by-step explanation:
1/3 x - 4x -5/3x + 1/3 x +3
2/3 x -4 -5/3x + 3
3/3x -4x +3
-1 1x -4x + 3
- 5x + 3
:)
<span>-6y + 14 + 4y = 32
Combine like terms:
-2y + 14 = 32
Take away 14 from both sides:
-2y = 18
Divide by -2 on both sides:
y = -9
Answer: y = -9 (Answer A)</span>
8
A=LW
72= 9(2x+4)
72= 18X+36
Sub 36
36= 18X
Divide by 18
X=2
Length (2(2)+4)=8
9514 1404 393
Answer:
a (3,-1)
Step-by-step explanation:
The number that "completes the square" is the square of half the x-coefficient, (-6/2)^2 = 9. Rearranging the given function to include the square trinomial, we have ...
f(x) = x^2 -6x +9 -1 . . . . . . . here, we have 8 = 9 - 1
f(x) = (x -3)^2 -1 . . . . . . . . . . vertex form
Comparing this to the generic vertex form ...
f(x) = (x -h)^2 +k . . . . . . . vertex at (h, k)
we see that h=3 and k=-1.
The vertex is (h, k) = (3, -1).
