Answer:
The limit of the function does not exists.
Step-by-step explanation:
From the graph it is noticed that the value of the function is 6 from all values of x which are less than 2. At x=2, the line y=6 has open circle. It means x=2 is not included.
For x<2
The value of the function is -3 from all values of x which are greater than 2. At x=2, the line y=-3 has open circle. It means x=2 is not included.
For x>2
The value of y is 1 at x=2, because of he close circles on (2,1).
For x=2
Therefore the graph represents a piecewise function, which is defined as
The limit of a function exist at a point a if the left hand limit and right hand limit are equal.
The function is broken at x=2, therefore we have to find the left and right hand limit at x=2.
Since the left hand limit and right hand limit are not equal therefore the limit of the function does not exists.
If your quadratic function has x intercepts of 4 and -3, that means that the points are (4, 0) and (-3,0), or the values x has when y=0. Since your quadratic (which represents y) must then equal 0, then your factors of that equation would potentially equal 0 as well. That means that if x=4, then subtract 4 from each side to get x-4=0, and so one factor is (x-4) Similarly your other factor would be (x+3).
<span>To get your quadratic, FOIL: </span>
<span>(x-4)(x+3)=0 </span>
<span>x^2-x-12=0</span>
You would use the formula c=<span>πd</span>
Which tables since there’s no picture
We can solve this by substitution. Let's make the second equation equal y by adding y to both sides. The equation would look like:
x-5=y
Let's plug that equation in for y in the first equation.
2x+x-5-10=0
Let's add like terms.
2x+x-15=0
Let's add 15 to both sides.
2x+x=15
3x=15
Now divide.
15÷3=5=x
Now, x equals 5, so that means y equals 0. Let's check in both equations.
2(5)+0-10=0
10+0-10=0
0=0
5-0-5=0
0=0
So, the solution of the system shown is (5,0). I also included a graph so you could see where they intercept. The solution, when put in the graph, is the x intercept.
