The graph represents function 1, and the equation represents function 2:
2 answers:
Answer:
Option D.
Step-by-step explanation:
The slope of a horizontal line is 0.
It is given that the function 1 is a horizontal line that passing through the y-axis at y = 4.
It means the rate of change of function 1 is 0.
The slope intercept form of a linear function is
.... (1)
where, m is slope and b is y-intercept.
The function 2 is
.... (2)
On comparing (1) and (2), we get
The rate of change of function 2 is 8.
The difference between rate of change is
The rate of change of function 2 is 8 more than the rate of change of function 1.
Therefore, the correct option is D.
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Let's see what to do buddy...
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To do this, we have to multiply the face and denominator of the fraction by the denominator conjunction, which is :
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
Reminder :
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
So we have :
The other is in the photo.
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And we're done.
Thanks for watching buddy good luck.
♥️♥️♥️♥️♥️
Answer:
Step-by-step explanation:
The domain of all polynomials is all real numbers. To find the range, let's solve that quadratic for its vertex. We will do this by completing the square. To begin, set the quadratic equal to 0 and then move the -10 over by addition. The first rule is that the leading coefficient has to be a 1; ours is a 2 so we factor it out. That gives us:
The second rule is to take half the linear term, square it, and add it to both sides. Our linear term is 2 (from the -2x). Half of 2 is 1, and 1 squared is 1. So we add 1 into the parenthesis on the left. BUT we cannot ignore the 2 sitting out front of the parenthesis. It is a multiplier. That means that we didn't just add in a 1, we added in a 2 * 1 = 2. So we add 2 to the right as well, giving us now:
The reason we complete the square (other than as a means of factoring) is to get a quadratic into vertex form. Completing the square gives us a perfect square binomial on the left.
and on the right we will just add 10 and 2:
Now we move the 12 back over by subtracting and set the quadratic back to equal y:
From this vertex form we can see that the vertex of the parabola sits at (1,-12). This tells us that the absolute lowest point of the parabola (since it is positive it opens upwards) is -12. Therefore, the range is R={y|y ≥ -12}