Explanation for Question 34 From the Math (Calc) Section on the Official Sat Practice Test 6
Question 34 reads and the X, 2 Y plane, the graph of Y is equal to three X squared, 3 minus 14 X intersects the graph of Y equals X at the points 4 zero, zero, and a. 5 What is the value of a, 6 okay, so let's kind of try and picture this in our heads. 7 So we've got the graph of Y is equal to X, 8 which is going to look approximately like this. 9 It crosses the origin and it has a slope 10 of one. And for this graph, 11 we know that it obviously also hits the origin because it says that it 12 intersects at zero zero. And because we see that there is no constant term 13 on the end, and we know that it curves upward because it has a 14 positive coefficient in front of the X squared. 15 So it's going to look something like this, 16 and we can see that there's an intersection that happens here as they say 17 that there's also an intersection somewhere at a a, 18 so let's set these two equations equal to each other to find the solution, 19 because that's telling us where, when they're equal to each other, 20 the Y is equal to the Y and the X is equal to the 21 X. 22 So let's say three X squared minus 14, 23 X is equal to X. 24 Now, one thing that you immediately notice is that you can divide both sides 25 of the equation by X, and that should help simplify it. 26 So we're left with three X minus 14 is equal to 27 one. Now adding 14 to both sides, 28 we get the three X is equal to 15 and then dividing each side 29 by three, we get that X is equal to five. 30 So we know that the point that we're given is a comma a, 31 so the value of a is the same thing as X a is equal 32 to five.