First set up a linear equation and using the x and y values in the table see if it solves.
It doesn't solve so we know it isn't linear. ( I won't show all those steps because they aren't needed.)
Using the quadratic formula y = ax^2 +bx +c
Build a set of 3 equations from the table:
C is the Y intercept ( when X is 0), this is shown in the table as 6
Now we have y = ax^2 + bx + 6
-2.4 =4a-2b +6
1.4 = a-b +6
Rewrite the equations
a=b/2 -2.1
1.4 = b/2-2.1 +6
b = 5
a = 5/2 -2.1 = 0.4
replace the letters to get y = 0.4x^2 + 5x +6
Answer:
2
Step-by-step explanation:
This implies
x+2=4
and
-(x+2)=-4.
x+2=4 implies x=2 since subtract 2 on both sides gives us x=2.
Solving -(x+2)=-4 should give us the same value.
Multiply both sides by -1:
x+2=4
It is the same equation as the other.
You will get x=2 either way.
Let's check:
Put both sides into your calculator and see if you get the same thing on both sides:
Left hand side gives 256/81.
Right hand side gives 256/81.
Both side are indeed the same for x=2.
1. Answer: Vertical shift up 3 units and vertical stretch by factor of 2
<u>Step-by-step explanation:</u>
f(x) = √x
g(x) = 2√x + 3
- adding 3 is a vertical shift up 3 units
- multiplying by 2 is a vertical stretch by factor of 2
2. Answer: Domain: [0, ∞)
Range: [3, ∞)
<u>Step-by-step explanation:</u>
g(x) = 2√x + 3
Domain: The restriction on "x" is that the radical must be greater than or equal to 0. So, x ≥ 0 Interval Notation: [0, ∞)
Range: Since the radical must be greater than or equal to 0, then 2√x is also greater than or equal to 0. Add 3 to that and y ≥ 3. Interval Notation: [3, ∞)
