For the first graph, part a). The limit of f(x) as x approaches 2 from the LEFT follows the line y = 4, which ends at the hollow point (2, 4). This means that the limit is 4.
First graph, part b), The limit as x approaches 2 from the RIGHT follows y = -1, with a hollow point at (2, -1). This means that the limit is -1.
Second graph: We approach x = 3 from the left, following the downward sloping line which ends at the hollow point (3, -1). This means that the limit is -1.
Third graph: We approach x = 3 from the right, following the horizontal line that ends at (3, 3). This means that the limit is 3.
Note that when talking about one-sided limits, we use the hollow point's y-value which the function approaches. However, when looking at the value of f(x), we would use the solid point.
Answer:
20-25
Step-by-step explanation:
i think but not sure
To apply the changes to the equation of a vertical stretch of 4 and a translation of 3 units to the right, as well as the correct answer would be choice B.
The reason for this is when you apply a vertical stretch, because it changes the y-values (which causes it to vertically stretch or appear skinnier when graphed), you would multiply 4 to f(x) which would look like 4x^2.
Then, since you have a reflection over the x-axis, you must multiply a -1 to f(x) to reflect it over the x-axis which would result in -4x^2.
Finally, it also asks to shift the graph right 3 which by moving it right, you change the x values meaning you will perform f(x-3) to achieve this (subtract the value from x when you move right, and add the value to x when you move left).
This therefore results in your answer, the new graph would be
g(x)= -4(x-3)^2 or choice B.
Step-by-step answer:
Given:
k(x)=8x+7q...........................(1)
Evaluate: k(2q^2+3q) .......(2)
Solution:
From (1)
k(x) = 8x+7q
Substitute x=2q^2+3q into (1)
k(2q^2+3q)
= 8(2q^2+3q) + 7q
= 16q^2 + 24q + 7q
= 16q^2 + 31q
