Answers7, 11 are linear.
9 is a quadratic. y = 2x^2 - 3
Problem 11First find the slope. Use the first two points to do that.
y2 = 8
y1 = 6
x2 = - 3
x1 = - 1
<u>Sub and solve</u>
y = (y2 - y1) / (x2 - x1)
y = (8 - 6) / (-3 - - 1)
y = 2/- 2
y = - 1
<u>Step 2</u>
Find the y intercept
So far we have
y = - x + b
x = 0
y = 5
5 = 0 *-1 + b
b = 5
<u>Answer</u>
y = -x + 5
Problem 9 is the quadratic.
y = a*x^2 + b
when x = 0
y = - 3
y = ax^2 - 3 Now use any other point to solve for a.
Use (2,5)
5 = a(2)^2 - 3 Add 3 to both sides.
5 + 3 = a*4
8 = 4a Divide by 4
8/4 = a
2 = a
<u>check</u>
Use (-1,1) to check
y = 2x^2 - 3
y = 2(-1)^2 - 3
y = 2*1 - 3
y = 2 - 3
y = - 1 Which is as it should be.
Problem 7 Is linear. Do the same way as 11.
The graph is included below.
See if you can get the equation line.
You should get
y = 2x - 4
