Math: fc3d783a

ID: fc3d783a

(SAT Suite Question Bank > Find Questions > Assessment: SAT + Test: Math + Domain: Advanced Math)

Comment: Figure out the line and the rest will be easy.

Method 1: Any equation in which y equals a number, without any other unknowns, will produce a horizontal line in a coordinate plane. If such a line intersects a parabola at exactly one point, then that point must be the vertex (or else the line would intersect the parabola two times). We can figure out from 2y = 4.5 that the vertex will have a y-value of 2.25. This information will come in handy in a moment.

To find the x-value of the vertex of a parabola in standard form—y = ax^2 + bx + c—use the formula        -(b/2a). Using the given parabola,

x-vertex = -((b)/2(-4))

x-vertex = -(-(b/8))

x-vertex = b/8

Now, we can use our earlier y-value for the vertex to solve for b.

(2.25) = -4(b/8)^2 + (b)(b/8)

2.25 = -4(b^2/64) + b^2/8

2.25 = -b^2/16 + b^2/8

36 = -b^2 + 2b^2

36 = b^2

6 = b

The correct answer is 6.

Method 2: This question is no match for Desmos. Type in the two equations and add a slider for b. Since b is a positive constant, you can test a few positive values and adjust the range for b as necessary.

As you move the slider to the right, the parabola jumps up and to the right. When b = 6, the parabola and the line touch, as shown. The correct answer is 6.

Method 3: If you wanted to work through the system of equations, you could do so through substitution.

2y = 4.5 (y = 2.25) and y = -4x^2 + bx

(2.25) = -4x^2 + bx

4x^2 - bx + 2.25 = 0

This next part is tricky, but what determines the number of solutions of a quadratic function is the discriminant, the part of the quadratic formula in the numerator that is under the square root, b^2 - 4ac.

- If b^2 - 4ac is positive, there will be two real solutions

- If b^2 - 4ac is 0, there will be one real solution (i.e. the quadratic function is a perfect square)

- If b^2 - 4ac is negative, there will be no real solutions

All the other components of the quadratic formula help to determine the exact values of the solution or solutions, but not how many solutions the function may have. You can thus ignore everything but the part that is pertinent to answering this question. Since the setup tells you that the two graphs intersect at exactly one point, there must be one solution, and you can use the middle formula above. And remember, these a, b, and c values all come from the standard form of a parabola, discussed above in method 1.

4x^2 - bx + 2.25 = 0

b^2 - 4ac = 0

(-b)^2 - 4(4)(2.25) = 0

b^2 - 36 = 0

b^2 = 36

b = 6

The correct answer is 6.

This is the most labor-intensive method of the three shown above, and I would therefore recommend it only if the other two methods eluded you. Even so, knowing how to derive the x-value of the vertex or the number of solutions from the standard form of a parabola is important if you want to earn a top score in Math.