<p>In year y, the value in dollars,v, of a certain painting created in 1970 is given by the equation
v=q+(r-6(y-1970))^2, where q and r are constants. If the painting reached its lowest value in 1990, when it was worth $500, what was the painting’s value,in dollars, in the year 2000?</p>
<p>I can’t seem to understand how to do this problem without having to use advanced math concepts that aren’t tested on the SAT. So is there an easier way of doing this problem???</p>
<p>wow. that is difficult…10char</p>
<p>Is the answer 4100?</p>
<p>(Want to be sure I am right before I try to explain it)</p>
<p>yes the answer is 4100!!!</p>
<p>Victory!
Here is how I did it:
First, make sure you picture it as a parabola. It has a squared term, and it tells you THE minimum. Thus, it is a parabola.</p>
<p>First, you can figure q out by that minimum alone. q is the vertical shift of the equation. At the minimum of a parabola, the Y is the vertical shift. Given the min: (1990, 500) you get the vertical shift: 500.</p>
<p>(I hope you have done transformations in your math class, or this would be rather confusing… Getting that q is essential to figuring out how to solve it)</p>
<p>So q = 500. From there, you have 1 variable (r) and can solve for r algebraically. </p>
<p>v=q+(r-6(y-1970))^2</p>
<p>Using the min: </p>
<p>500 + (r-6(1990-1970))^2 = 500
(r-6(20))^2 = 0
(r - 120) ^ 2 = 0
(r-120)(r-120) = 0
Thus, r -120 = 0
And thus… r = 120</p>
<p>Then with q = 500 and r = 120 you just plug them into the equation with 2000 as the year.
That is:
v = 500 + (120-6(2000-1970))^2 = 4100</p>
<p>I hope that makes sense. It was kind of hard to explain, maybe somebody else will be able to do it more clearly.</p>
<p>good job. that was a hard one.</p>