7y^2 + 10y + 3
7y^2 + 7y + 3y + 3
(7y^2+7y) + (3y+3)
7y(y+1) + 3(y+1)
(7y+3)(y+1)
So the final answer is <span>(7y+3)(y+1)</span>
Check the picture below.
make sure your calculator is in Degree mode.
The range is the value of y. We know that Ix-12I is always ≥0, no matter what x is, so Ix-12I-2 is always ≥ -2, the answer is B.
Evaluate the function
g(x) = 2x2 + 3x – 5 for the input values -2, 0, and 3.
This is tedious math work but necessary to sharpen your skills.
Let x = -2
g(-2) = 2(-2)^2 + 3(-2) – 5
g(-2) = 2(4) - 6 - 5
g(-2) = 8 - 11
g(-2) = -3
Now let x = 0 and repeat the process.
g(0) = 2(0)^2 + 3(0) - 5
g(0) = 0 + 0 - 5
g(0) = -5
Lastly, let x = 3.
g(3) = 2(3)^2 + 3(3) - 5
g(3) = 2(9) + 9 - 5
g(3) = 18 + 9 - 5
g(3) = 27 - 5
g(3) = 22
Did you follow through each step?
