The vertex form of the equation of a parabola is $y=4(x-2)^2-1$. What is the standard form of the equation?
A. $y=4 x^2-3$
B. $y=4 x^2-16 x+2$
C. $y=16 x^2-5$
D. $y=4 x^2 - 16 x+15$
Answer (2)
Expand the squared term: ( x − 2 ) 2 = x 2 − 4 x + 4 .
Substitute back into the equation: y = 4 ( x 2 − 4 x + 4 ) − 1 .
Distribute the constant: y = 4 x 2 − 16 x + 16 − 1 .
Simplify to standard form: y = 4 x 2 − 16 x + 15 . The answer is y = 4 x 2 − 16 x + 15 .
Explanation
Understanding the Problem The problem gives us the vertex form of a parabola's equation, which is y = 4 ( x − 2 ) 2 − 1 , and asks us to find the standard form of the equation.
Expanding the Squared Term To convert from vertex form to standard form, we need to expand and simplify the given equation. First, let's expand the squared term ( x − 2 ) 2 . Recall that ( a − b ) 2 = a 2 − 2 ab + b 2 . Therefore, ( x − 2 ) 2 = x 2 − 2 ( x ) ( 2 ) + 2 2 = x 2 − 4 x + 4
Substituting Back into the Equation Now, substitute this back into the original equation: y = 4 ( x 2 − 4 x + 4 ) − 1
Distributing the Constant Next, distribute the 4 across the terms inside the parentheses: y = 4 x 2 − 16 x + 16 − 1
Simplifying to Standard Form Finally, combine the constant terms: y = 4 x 2 − 16 x + 15
Identifying the Correct Option The standard form of the equation is y = 4 x 2 − 16 x + 15 . Comparing this to the given options, we see that it matches option D.
Examples
Understanding quadratic equations is crucial in various real-world applications. For instance, engineers use parabolas to design suspension bridges and antennas. Architects apply quadratic functions to model the shape of arches and other structures. In physics, projectile motion follows a parabolic path, making quadratic equations essential for calculating trajectories.
The standard form of the equation y = 4 ( x − 2 ) 2 − 1 is y = 4 x 2 − 16 x + 15 . The correct answer is option D. This form highlights the coefficients of the quadratic equation and is used in various applications in mathematics.
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