Can i have with this problem . algebra 1 quadratic functions

We can find the revenue equation with the data provided using the generic quadratic equation, ax^2+bx+c and the given points...

9a+3b+c=2530
4a+2b+c=2530
a+b+c=2520  getting differences

5a+b=0
3a+b=10  and again

2a=-10, a=-5 making 3a+b+10 become:
-15+b=10, b=25, making a+b+c=2520 become:
-5+25+c=2520, c=2500 so

r(x)=-5x^2+25x+2500

r(0)=2500 so the first statement is true.

dr/dx=-10x+25

dr/dx>0 when -10x+25>0, -10x>-25, x<2.5

This means that the second statement is only true when she decreases the price by more than $2.50.  If she decreases the price by less than $2.50, her revenue will increase...

r(1)=-5x^2+25x+2500=2520

So the third statement is true.

Maximum revenue will occur when dy/dx=0

dr/dx=-10x+25=0 only when -10x=-25, x=2.5

She will maximize revenue if she decreases the price by $2.50
( r(2.5)=2531.25 is maximum revenue for the curious :P)

So the fourth statement is false.

r(x)=-5x^2+25x+2500 so the decrease in price for 24 is 1 and for 21 is 4

r(1)=2520, r(4)=2520

So the fifth statement is true.