Explanation for Question 13 From the Math (No Calc) Section on the 2019 May Sat
Question 13 asks, which of the following is the graph in the X, 2 Y plane of the given equation. And they give us the equation, 3 Y equals negative X squared plus four X. 4 So the first step here is to plug in zero for both Y and 5 X, to see what the Y intercepts and the X intercepts are. 6 So let's start by plugging in zero for X to find the Y intercept. 7 When we do that, we get Y is equal to zero 8 plus zero. So Y is equal to zero, 9 which means that this graph should have one Y intercept at zero. 10 From that information we can cross off a and B because 11 a has a negative y-intercept negative four, 12 and B has positive four as it's y-intercept. 13 After that, we can plug in zero for Y to solve for the X 14 intercepts. So let's do that. 15 Let's say that Y is equal to zero. 16 So then we have zero is equal to negative X squared, 17 plus four X. 18 We know that the first zero is X because we can factor X out 19 negative X plus four. 20 We can also divide this by negative one to make this X minus four. 21 So we know that the second X intercept should be at X is equal 22 to four. So X is equal to zero and X is equal to four. 23 That means that D is the answer because it has both of those X 24 intercepts, whereas C has the X intercept, 25 negative four and zero.