Question

the graph shows the daily revenue earned selling coupon cards in a fundraiser if the price is decreased by x dollars. which function can be used to determine the daily revenue if the price of a coupon card is decreased by x dollars? r(x)= -x² + 6x + 40 r(x)=-(x - 4)(x + 10) r(x)=-(x² - 14x + 40) r(x)=-(x + 3)² + 49

Explanation:

Step1: Recall the general form of a quadratic function

The general form of a quadratic function is $y = ax^{2}+bx + c$. We can test the given points $(0,40),(3,49)$ in each of the functions.

Step2: Test the point $(0,40)$

For $R(x)=-x^{2}+6x + 40$, when $x = 0$, $R(0)=-(0)^{2}+6(0)+40=40$.
For $R(x)=-(x - 4)(x + 10)=-(x^{2}+10x-4x - 40)=-x^{2}-6x + 40$, when $x = 0$, $R(0)=-(0)^{2}-6(0)+40=40$.
For $R(x)=-(x^{2}-14x + 40)=-x^{2}+14x - 40$, when $x = 0$, $R(0)=-(0)^{2}+14(0)-40=-40$ (rejected).
For $R(x)=-(x + 3)^{2}+49=-(x^{2}+6x + 9)+49=-x^{2}-6x+40$, when $x = 0$, $R(0)=-(0)^{2}-6(0)+40=40$.

Step3: Test the point $(3,49)$

For $R(x)=-x^{2}+6x + 40$, when $x = 3$, $R(3)=-(3)^{2}+6\times3 + 40=-9 + 18+40=49$.
For $R(x)=-(x - 4)(x + 10)=-x^{2}-6x + 40$, when $x = 3$, $R(3)=-(3)^{2}-6\times3 + 40=-9-18 + 40=13$ (rejected).
For $R(x)=-(x + 3)^{2}+49=-x^{2}-6x + 40$, when $x = 3$, $R(3)=-(3)^{2}-6\times3+40=-9-18 + 40=13$ (rejected).

Answer:

$R(x)=-x^{2}+6x + 40$