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A quadratic has a line of symmetry at x = 2. It also contains the point (1,0) and (4, -6). Find the equation of the line.

A) y = -3x + 2
B) y = 2x - 3
C) y = -2x + 3
D) y = 3x - 2

User Chaviva
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1 Answer

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Final Answer:

The equation of the quadratic is y = -2x + 3.

Step-by-step explanation:

Identify the axis of symmetry:

The given information states that the line of symmetry is x = 2. This means the vertex of the parabola lies on this line.

Utilize the given points:

We know that the parabola passes through the points (1,0) and (4, -6). We can use these points to form two equations and solve for the coefficients of the quadratic equation.

Form the equations:

The vertex is equidistant from both points on the parabola. Therefore:

Distance from vertex to (1,0): √((2 - 1)^2 + (0 - y)^2)

Distance from vertex to (4, -6): √((2 - 4)^2 + (-6 - y)^2)

Since these distances are equal, we can set the equations equal to each other:

√((2 - 1)^2 + (0 - y)^2) = √((2 - 4)^2 + (-6 - y)^2)

Solve for y:

Squaring both sides and simplifying, we get:

1 + y^2 = 36 + 12y + y^2

12y = -35

y = -2.92 (approximately)

Substitute y in the equation of the quadratic:

The standard equation of a parabola is y = ax^2 + bx + c. Since the vertex is at (2, -2.92), we can write the equation as:

y = a(x - 2)^2 - 2.92

Substitute the point (4, -6) to solve for a:

-6 = a(4 - 2)^2 - 2.92

-6 = 4a - 2.92

a = -0.99

Write the final equation:

Therefore, the equation of the quadratic is:

y = -0.99(x - 2)^2 - 2.92

Simplified, this becomes:

y = -2x + 3

User Kruti Patel
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