For example, how would your graph of f(x) = 2`x^{2}` be different from the graph of f(x) = .3`x^{2}`?
Or how is the graph of f(x) = 1.5 `x^{2}` different from the graph of f(x) = -1.5 `x^{2}`?
How did changing "a" in your equation (y = a`x^{2}`) affect your graph?

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

one way is to plot 2 points and connect them which is what the answer choices seem to be doing

it seems to be using the y intercept and the slope

y=mx+b, m=slope, b=y-intercept (where the line hits the y axis)

slope=rise/run

y-intercept is where the line croses the y axis which is the y value where x=0

you seem to have given the wrong slope (you didn't put a number in front of te x, it should be 2/3 or 3/2 looking at the answer options)

y=? x-1

m=? because of your fault

b=-1

so it has a slope of ? and y intercept of -1, meaning it pases through (0,-1)

so draw point (0,-1) (so answer is one of the first 2 option)

if slope is 2/3, then go up 2 and right 3 then draw line through those 2 points

if slope is 3/2, then go up 3 and right 2 then draw line through those 2 points

the answer is 9 because the question is telling how much it be together.