In the xy-plane, the vertex of the parabola y = x^2 + bx + c is V_1, and the vertex of the… — SAT

In the $xy$ -plane, the vertex of the parabola $y = x^2 + bx + c$ is $V_1$ , and the vertex of the parabola $y = -x^2 + dx + e$ is $V_2$ , where $b$ , $c$ , $d$ , and $e$ are constants. The parabola $y = x^2 + bx + c$ passes through $V_2$ , and the parabola $y = -x^2 + dx + e$ passes through $V_1$ . If the distance between $V_1$ and $V_2$ is $2\sqrt{5}$ , and the line passing through $V_1$ and $V_2$ has a positive slope, what is the value of $b + d$ ?

A
2
4Answer
C
8
D
16

Answer

The correct answer is 4. By writing the equations of the parabolas in vertex form,

y = (x - h_1)^2 + k_1

and

y = -(x - h_2)^2 + k_2

, and using the fact that each parabola passes through the other's vertex, we establish that the vertical difference between the vertices is equal to the square of their horizontal difference:

k_2 - k_1 = (h_2 - h_1)^2

. Substituting this into the distance formula

(h_2 - h_1)^2 + (k_2 - k_1)^2 = 20

yields a quadratic equation in terms of

(h_2 - h_1)^2

, which solves to

(h_2 - h_1)^2 = 4

. Since the slope of the line segment connecting the vertices is positive, the horizontal difference

h_2 - h_1

must be positive, which gives

h_2 - h_1 = 2

. Finally, expressing the linear coefficients in terms of the vertex coordinates using the vertex formula gives

b = -2h_1

and

d = 2h_2

, so

b + d = 2(h_2 - h_1) = 2(2) = 4

Step-by-Step Solution

Write each parabola in vertex form using the vertices

V_1(h_1, k_1)

and

V_2(h_2, k_2)

f(x) = (x - h_1)^2 + k_1

and

g(x) = -(x - h_2)^2 + k_2

Expressing the functions in vertex form allows us to easily use the condition that each parabola passes through the other's vertex.

Substitute the vertex

V_2(h_2, k_2)

into

f(x)

and

V_1(h_1, k_1)

into

g(x)

k_2 = (h_2 - h_1)^2 + k_1

and

k_1 = -(h_1 - h_2)^2 + k_2

, both of which simplify to

k_2 - k_1 = (h_2 - h_1)^2

Since the points lie on the respective curves, their coordinates must satisfy the equations, establishing a relation between horizontal difference

\Delta h = h_2 - h_1

and vertical difference

\Delta k = k_2 - k_1

Use the distance formula for

V_1

and

V_2

with the given distance

2\sqrt{5}

and substitute

\Delta k = (\Delta h)^2

(\Delta h)^2 + ((\Delta h)^2)^2 = (2\sqrt{5})^2 \implies (\Delta h)^4 + (\Delta h)^2 - 20 = 0

. Solving this quadratic in terms of

(\Delta h)^2

yields

(\Delta h)^2 = 4

The distance formula connects the horizontal and vertical differences, which we can solve using our relation from the previous step.

Determine the horizontal difference

\Delta h

using the slope condition.

Since the slope

m = \frac{\Delta k}{\Delta h}

is positive and we know

\Delta k = 4 > 0

, we have

\Delta h = h_2 - h_1 = 2

The positive slope restriction determines that the horizontal translation is to the right from

V_1

V_2

Relate the standard form coefficients

b

and

d

to the vertex coordinates to find

b + d

For

f(x) = x^2 + bx + c

, the vertex x-coordinate is

h_1 = -\frac{b}{2} \implies b = -2h_1

. For

g(x) = -x^2 + dx + e

, the vertex x-coordinate is

h_2 = \frac{d}{2} \implies d = 2h_2

. Thus,

b + d = -2h_1 + 2h_2 = 2(h_2 - h_1) = 2(2) = 4

The vertex x-coordinate formula

h = -\frac{\text{coefficient of } x}{2 \times \text{coefficient of } x^2}

connects the standard form coefficients back to

h_1

and

h_2

Key Concept

Identifying vertex coordinates of quadratic functions and applying translations and the distance formula in the coordinate plane.

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