This is an explanation of Bluebook test 4 Section 2b HARD question 22. BB4.2b.22 In the xy-plane, a parabola has vertex (9,-14) and intersects the x-axis at two points. If the equation of the parabola is written in the form y-ax2+bx+c, where a, b, and c are constants, which of the following could be the value of a+b+c? There are a few ways to answer this question, but this way made the most sense to me. To answer this question, you need to know at least two of the three forms of the quadratic equation AND be able to expand and condense in order to switch between different forms. We are given that the vertex is Positive 9 COMMA negative 14. First fill in your vertex in vertex form. Then expand. I like to write the equation in desmos to ensure I haven’t goofed somewhere. I write my original equation as f(x) = Then I use g(x) for the expanded version. They perfectly overlap, so I know I didn’t make any errors. I work toward getting my equation in standard form in order to get those a,b, and c values. Now I want to get all of my terms expressed in terms of a common variable. A is a B is -18a C is 81a-14 a+b+c = a -18a + 81a -14 Combining like terms we get 64a - 14 Now we turn to our answer choices. Let’s start by setting 64a-14 equal to answer choice A: 23 We find that a would be a negative number, and that doesn’t not make sense because our parabola opens upward. We know that because our vertex was positive 9 COMMA negative 14 and our parabola intersected the x-axis at two points. Let’s try answer choice D. We find that a is a positive number. It is the only positive a value we can calculate with the offered answer choices. Therefore D is our answer. Note that I didn't use the real numbers for the thumbnail because I didn't want to spoil that step. Disclaimer: SAT® is a trademark registered by the College Board, which is not affiliated with, and does not endorse this product. Disclaimer: Desmos is not affiliated with, and does not endorse this product.