The correct excluded value is 2.
You haven't shared the given line, so all I can do here is to invent a line and then show you how to write the equation of a new line which is parallel to mine and which has an x-intercept of 4.
My invented line: y = (2/3)x + 3
The new line MUST have the same slope: m = 2/3.
Then y = mx + b becomes y = (2/3)x + b. Find the x-intercept by setting y = 0 and solving for x: (2/3)x = 0 - b. Now replace x with 4 and find b:
-b = (2/3)(4) = 8/3. Then b = -8/3, and the new line is
y = (2/3)x - 8/3.
I think it’s (-10, -5) if your looking at the Xmin and Ymin
Add 1
0.7+1=1.7
1.7+1= 2.7
etc., etc.
Nice, already in vertex form
y=a(x-h)^2+k
(h,k) is vertex
therfor since (-3,6) is vertex
we are looking for something like
y=a(x-(-3))^2+6 simplified to
y=a(x+3)^2+6
A is ansre
