Answer:
vertex is (4,-4) and another point is (6,0) or you could use (2,0) or many other options :)
Step-by-step explanation:
The cool thing about this question your quadratic is in factored form so your x-intercepts are easy to figure out, they are 2 and 6.
So you can plot (6,0) and (2,0).
The vertex will lie half between x=2 and x=6... so it lays at (6+2)/2=4
We just have to find the y-coordinate for when x=4.
Plug in 4 gives you (4-2)(4-6)=(2)(-2)=-4.
So the vertex is at (4,-4).
Answer:
<-|---------------|----------------|--------------|->
0 1/3 2/3 1
Step-by-step explanation:
easy peasy
Answer:
the parabola can be written as:
f(x) = y = a*x^2 + b*x + c
first step.
find the vertex at:
x = -b/2a
the vertex will be the point (-b/2a, f(-b/2a))
now, if a is positive, then the arms of the parabola go up, if a is negative, the arms of the parabola go down.
The next step is to see if we have real roots by using the Bhaskara's equation:

Now, draw the vertex, after that draw the values of the roots in the x-axis, and now conect the points with the general draw of the parabola.
If you do not have any real roots, you can feed into the parabola some different values of x around the vertex
for example at:
x = (-b/2a) + 1 and x = (-b/2a) - 1
those two values should give the same value of y, and now you can connect the vertex with those two points.
If you want a more exact drawing, you can add more points (like x = (-b/2a) + 3 and x = (-b/2a) - 3) and connect them, as more points you add, the best sketch you will have.
Do you mean he had 3 shelves and 5 on each? If so the answer is 15