In the xy-plane, the graph of the quadratic function f(x) = ax^2 + bx + c, where a > 0, ha… — SAT

In the $xy$ -plane, the graph of the quadratic function $f(x) = ax^2 + bx + c$ , where $a > 0$ , has vertex $V$ . The graph intersects the $x$ -axis at points $A$ and $B$ . If triangle $VAB$ is an equilateral triangle with an area of $12\sqrt{3}$ , what is the value of $a$ ?

A
$\frac{1}{8}$
$\frac{1}{2}$ Answer
C
$1$
D
$2$

Answer

The correct answer is the value 1/2.

The value 1/2 is correct because the area of the equilateral triangle determines its side length to be

4\sqrt{3}

and its height to be

6

. Since the parabola opens upwards and its base lies on the

x

-axis, the vertex is

V(h, -6)

. The two

x

-intercepts are symmetric about the axis of symmetry

x = h

, meaning they are located at

h \pm 2\sqrt{3}

. Substituting one of these points into the vertex form

f(x) = a(x-h)^2 - 6

yields

0 = 12a - 6

, which gives

a = 1/2

Step-by-Step Solution

Determine the side length and height of the equilateral triangle from the given area.

The side length

s

4\sqrt{3}

, and the height of the triangle is

6

The area of an equilateral triangle with side length

s

is given by the formula

\text{Area} = \frac{s^2\sqrt{3}}{4}

. Setting this equal to

12\sqrt{3}

yields

s^2 = 48

, which means

s = \sqrt{48} = 4\sqrt{3}

. The height of the triangle is given by

h_{\text{tri}} = \frac{s\sqrt{3}}{2} = \frac{4\sqrt{3} \cdot \sqrt{3}}{2} = 6

Relate the height and side length of the triangle to the vertex and

x

-intercepts of the parabola.

The vertex of the parabola is

V(h, -6)

and the

x

-intercepts are at

x = h \pm 2\sqrt{3}

Because

a > 0

, the parabola opens upwards, meaning the vertex

V(h, k)

lies below the

x

-axis. Since the base of the triangle lies on the

x

-axis, the vertical distance from the vertex to the

x

-axis is the height of the triangle, so

k = -6

. The axis of symmetry

x = h

bisects the base

AB

, so the

x

-intercepts are located at a distance of half the side length,

\frac{s}{2} = 2\sqrt{3}

, to the left and right of

h

Write the quadratic function in vertex form and substitute one of the

x

-intercepts to solve for

a

a = \frac{1}{2}

The vertex form of the quadratic function is

f(x) = a(x - h)^2 - 6

. Substituting the

x

-intercept

(h + 2\sqrt{3}, 0)

into this equation gives

0 = a(h + 2\sqrt{3} - h)^2 - 6

, which simplifies to

0 = a(2\sqrt{3})^2 - 6 \implies 12a = 6 \implies a = \frac{1}{2}

Key Concept

Using the symmetry properties and vertex form of a quadratic function to relate the geometric features of its graph to its algebraic coefficients.

Alternative Method

Alternatively, you can use the relationship between the roots and coefficients. For any quadratic equation

ax^2 + bx + c = 0

with real roots, the distance between the roots is

\frac{\sqrt{b^2 - 4ac}}{|a|}

. Since the triangle is equilateral, the height is

\frac{\sqrt{3}}{2}

times this distance. The vertex is at

(-\frac{b}{2a}, -\frac{b^2-4ac}{4a})

. Setting the absolute value of the vertex

y

-coordinate equal to the height allows you to solve for

a

directly when combined with the area formula.

Estimated Time:3m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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