Y = 2x-3
The second line will have its Y-Intercept at -3 and its slope will be up 2 and over 1 which will allow it to go through (4,5) and the line will be parallel to Y = 2x+2
For the first line we have a slope of (y2-y1)/(x2-x1)
(2--2)/(1--1)=4/2=2 so we have:
y=2x+b, now solve for b with either of the points, I'll use: (1,2)
2=2(1)+b
b=0 so the first line is:
y=2x
Now the second line:
(1-10)/(4--2)=-9/6=-3/2 so far then we have:
y=-3x/2+b, using point (4,1) we solve for b...
1=-3(4)/2+b
1=-6+b
b=7 so
y=-3x/2+7 or more neatly...
y=(-3x+14)/2
...
The solution occurs when both the x and y coordinates for each are equal, so we can say y=y, and use our two line equations...
2x=(-3x+14)/2
4x=-3x+14
7x=14
x=2, and using y=2x we see that:
y=2(2)=4, so the solution occurs at the point:
(2,4)
Answer:
8
Step-by-step explanation:
There are 2 roots so the only way to complete the square is,
![y=2x^2+8x-9\\y=2[(x^2+4x)]-9\\y=2[(x^2+4x+4)-4]-9\\y=2[(x+2)^2-4]-9\\y=2(x+2)^2-8-9\\y=2(x+2)^2-17](https://tex.z-dn.net/?f=y%3D2x%5E2%2B8x-9%5C%5Cy%3D2%5B%28x%5E2%2B4x%29%5D-9%5C%5Cy%3D2%5B%28x%5E2%2B4x%2B4%29-4%5D-9%5C%5Cy%3D2%5B%28x%2B2%29%5E2-4%5D-9%5C%5Cy%3D2%28x%2B2%29%5E2-8-9%5C%5Cy%3D2%28x%2B2%29%5E2-17)
Just factor 2 out of 2x^2+8x (just ignore the -9) then find the number that will make the terms be able to complete the square.
then complete the square and multiply 2 inside the brackets.
subtraction as you already get the vertex form and know how to complete the square.
Vertex Form: 
