Very simple.
Let's say you have an equation.
f(x) = x^2
You are asked to find the value for y when x equals 1.
The new equation is: f(1) = (1)^2
f(1) = 1
When x = 1, y = 1.
The same concept is applied here.
In the graph, where does x equal 0?
It equals zero at the origin.
Is there any y-value associated with 0?
Yes, there is.
Y equals five when x equals 0.
So
h(0) = 5
12 + 5x > 2(8x - 6) - 7x
12 + 5x > 16x - 12 - 7x
5x - 16x + 7x > -12 - 12
-4x > -24 / : (-4)
x < 6
Answer:
Step-by-step explanation:
1. Distribute the X across the equation: 
2.
3.
3.
To factor the given function, find a term that is divisible by both terms. For this problem, the factor would be 2, since 2x^2 and -18 are both divisible by 2.
2(x^2 - 9)
You could further factor out x^2 - 9 because it is factorable by (x - 3) and (x + 3). Thus, it would be
2(x -3)(x+3)
