|
Question 31562: Can you please help me solve this?
a) Find x and y intercepts of the parabola
y=x^2-6x+5
b) What is the Vertex of this parabola?
Answer by josmiceli(19441)   (Show Source):
You can put this solution on YOUR website! when x = 0
y = 5
thats the y-intercept
when y = 0
y=x^2-6x+5
0 = x^2 -6x + 5
0 = (x - 5)(x - 1)
this is true if x = 5 or x = 1
the x-intercepts are (5,0) and (1,0)
the vertex occurs where y(x + k) = y(x - k)
x is the vertex
(x+k)^2 -6(x+k) + 5 = (x-k)^2 -6(x-k) +5
x^2 +2kx +k^2 -6x -6k +5 = x^2 -2kx +k^2 -6x +6k +5
cancel the terms that are the same on both sides
2kx -6k = -2kx +6k
4kx = 12k
kx = 3k
x = 3
when x = 3
y = 3^2 -6(3) +5
y = -4
(3, -4) is the vertex
|
|
|
| |
