Where's the so-called "parabola tool?"
The vertex is relatively easy to find. Find the midpoint of the segment of the x-axis connecting x=2 and x=6; it is x=4. Then evaluate the function at x=4:
f(4) = (4-2)(4-6) = 2(-2) = -4. The vertex is at (4,-4).
Plot (4,-4). Then (arbitrarily) let x=0 and find y: f(0) = y = (-2)(-6) = -12.
So, when x = 0, y = -12.
Plot this point (1, -12). realize that the parabola is symmetric about the line x=4. (1,-12) is 3 units to the left of the axis of symm, and thus another point on the parab. is 3 units to the right of this axis, x= 7. Plot (7,-12). Now draw the parabola thru (7,-12), (1, -12) and (4, -4). The parab. opens up.
Answer: i got 5/4 i just looked it up so ion know how to do it
1. X= 61/8
2. X= 61/9
3. Y= 20/3
Answer:
Step-by-step explanation:
Problem One
19 + 2*ln(x) = 25 Subtract 19 from both sides
19 - 19 + 2ln(x) = 25 - 19 Combine
2* ln(x) = 6 Divide by 2
ln(x) = 3 Take the inverse log
e^(ln(x)) = e^3 Reduce
x = 20.09
Problem 2
Getting this to graph in Desmos can be a bit of trouble. The correct format is y = abs(x^2 - 3x + 1). Knowing about abs and getting the quadratic to work are two separate [and difficult] problems.
The graph for this is below.
The answers (according to Desmos) are (2,1) and (3.4, 2.4). I think you should include D but that is a matter of opinion. I think it's a typo. Whoever made the question up may actually get that answer.