The graph below shows a company's profit f(x), in dollars, depending on the price of erasers x, in dollars, being sold by the company:

graph of quadratic function f of x having x intercepts at ordered pairs 0, 0 and 8, 0. the vertex is at 4, 270

part a: what do the x-intercepts and maximum value of the graph represent? what are the intervals where the function is increasing and decreasing, and what do they represent about the sale and profit?
part b: what is an approximate average rate of change of the graph from x = 1 to x = 4, and what does this rate represent?

  a) the x-axis represents the price of erasers, as stated in your question. so the x-intercepts are when f(x) = 0; i.e. there is no profit and the company 'breaks even'. the maximum value is the optimum price to charge to maximise profits. since the function is increasing for 0 ≤ x < 4, increasing the price in that region will increase profits. similarly, since the function is decreasing for 4 < x ≤ 8, increasing the price in that region will reduce profits. if the price is more than $8, then there will be negative profit and the company will lose money.b) if the profit, which is f(x), was at least $120, then we're basically solving f(x) ≥ 120. looking at the graph, you can see when the price (x) is $1 or $7, exactly $120 is earned, and for prices between $1 and $7 the company earns more. so the domain will be 1 ≤ x ≤ 7. 
c) from x = 1 to x = 4, the approximate rate of change will be given by the distance formula (let 1,120 and 4,270 be your points). do the calculations and you should get 50 as your gradient. that means that for every dollar that the company increases their prices (in the region 1 ≤ x ≤ 4) they will get roughly $50 more in profits. 

2. a) h(t) = -16t^2 + vt + s 
but initially (at t = 0), the object is 90ft above the ground. so substitute this in. 
90 = -16(0) + v(0) + s 
90 = s 
also, from the question, v = 50 
so we can say that h(t) = -16t^2 + 50t + 90 
so to find the time when it hits the ground, we need to find when the height is 0ft above the ground 
in other words, we're solving -16t^2 + 50t + 90 = 0 
b) i'm going to assume you don't know calculus, since this is an algebra 1 question. however, we know that the vertex of a parabola is t = -b/2a (it's t, rather than x, since this parabola has t's in it). so let's sub that in 
t = -50 / -32 = 25/16 
at this point, the parabola will be at its highest point (since it's concave down). so that means we can find the maximum height by substituting this value of t into h(t) 
so h(t) = -16(25/16)^2 + 50(25/16) + 90 
h(t) = -625/16 + 1250/16 + 90 
h(t) = -625/16 + 1250/16 + 1440/16 
h(t) = 2065/16, or roughly 129ft above the ground 
c) to solve h(t) = g(t) using a graph, we need a table of values for each one 
for h(t) 
t 0 1 2 3 4 5 
h(t) 90 124 126 96 34 < 0 
and for g(t) 
t 0 1 2 3 4 5 
g(t) 28 76.8 125.6 174.4 223.2 272 
you can see that at t = 2, they are roughly equal 
so 2 seconds after release, the two objects are at roughly the same height in the sky, and if they were released from the same spot, they would be close to colliding