the graph shows the daily revenue earned selling coupon cards in a fundraiser if the price is decreased by x…
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
Answer
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 check each option by substituting the points from the graph $(x,y)$ into the functions.
Step2: Substitute the point $(0,40)$
When $x = 0$, for the function $R(x)=-x^{2}+6x + 40$, we have $R(0)=-(0)^{2}+6(0)+40=40$. For $R(x)=-(x - 4)(x + 10)=-(x^{2}+10x-4x - 40)=-x^{2}-6x + 40$, $R(0)=-0^{2}-6(0)+40 = 40$. For $R(x)=-(x^{2}-14x + 40)=-x^{2}+14x - 40$, $R(0)=-0^{2}+14(0)-40=-40$ (rejected). For $R(x)=-(x + 3)^{2}+49=-(x^{2}+6x + 9)+49=-x^{2}-6x+40$, $R(0)=-0^{2}-6(0)+40 = 40$.
Step3: Substitute the point $(1,45)$
For $R(x)=-x^{2}+6x + 40$, when $x = 1$, $R(1)=-(1)^{2}+6(1)+40=-1 + 6+40=45$. For $R(x)=-(x - 4)(x + 10)=-x^{2}-6x + 40$, when $x = 1$, $R(1)=-(1)^{2}-6(1)+40=-1-6 + 40=33$ (rejected). For $R(x)=-(x + 3)^{2}+49=-x^{2}-6x + 40$, when $x = 1$, $R(1)=-(1)^{2}-6(1)+40=-1-6 + 40=33$ (rejected).
Answer:
$R(x)=-x^{2}+6x + 40$