Quadratic Function In Vertex Form Zeros At -1 And 3 Point (4, 5)

To determine the equation representing a quadratic function in vertex form, given its zeros and a point it passes through, we'll embark on a step-by-step journey. Our specific challenge involves a quadratic function boasting zeros at -1 and 3, while gracefully traversing the point (4, 5). The quest culminates in pinpointing the correct vertex form equation from a quartet of options: a) $f(x) = 4(x-1)^2 + 3$, b) $f(x) = (x-1)^2 + 3$, c) $f(x) = (x-1)^2 - 4$, and d) $f(x) = (x-1)^2 - 3$.

Understanding Quadratic Functions and Their Forms

Before diving into the solution, let's solidify our understanding of quadratic functions and the significance of their various forms. A quadratic function, in its essence, is a polynomial function of degree two, generally expressed as $f(x) = ax^2 + bx + c$, where 'a', 'b', and 'c' are constants, and 'a' steadfastly refuses to be zero. The graph of such a function unfurls as a parabola, a symmetrical U-shaped curve that holds a wealth of information within its form.

The zeros of a quadratic function, also known as its roots or x-intercepts, are the points where the parabola intersects the x-axis. These are the x-values for which $f(x) = 0$. Knowing the zeros provides valuable anchors for constructing the function's equation.

The vertex form of a quadratic function, represented as $f(x) = a(x-h)^2 + k$, offers a particularly insightful perspective. In this form, (h, k) elegantly reveals the vertex of the parabola, the point where it reaches its minimum (if a > 0) or maximum (if a < 0) value. The vertex acts as the parabola's turning point and a cornerstone of its symmetry. The 'a' in vertex form mirrors the 'a' in the standard form, dictating the parabola's direction (upward if positive, downward if negative) and its vertical stretch or compression. The vertex form is highly prized for its ability to immediately spotlight the vertex, a key feature in various applications.

Leveraging Zeros to Construct the Quadratic Function

Our quadratic function, with zeros at -1 and 3, gifts us a foundational piece of its structure. We can express the function in the factored form: $f(x) = a(x - r_1)(x - r_2)$, where $r_1$ and $r_2$ are the zeros. Substituting our zeros, we get $f(x) = a(x - (-1))(x - 3)$, which simplifies to $f(x) = a(x + 1)(x - 3)$. This form elegantly captures the function's roots, but the 'a' remains an enigma, and the vertex is still veiled.

The point (4, 5), through which the function passes, becomes our next guiding star. This point assures us that when x = 4, f(x) gracefully equals 5. By substituting these values into our factored form, we can solve for 'a':

$5 = a(4 + 1)(4 - 3)$ $5 = a(5)(1)$ $5 = 5a$ $a = 1$

With 'a' unveiled as 1, our function takes a more definitive shape: $f(x) = (x + 1)(x - 3)$. While this factored form is insightful, our quest lies in the vertex form. Thus, we must embark on a transformation.

Transitioning from Factored Form to Vertex Form

To transmute our function from factored form to vertex form, we first expand the factored form: $f(x) = (x + 1)(x - 3) = x^2 - 3x + x - 3 = x^2 - 2x - 3$. This unveils the standard form of our quadratic function. Now, the magic of completing the square will usher us into vertex form.

Completing the square involves rewriting the quadratic expression in a way that reveals a perfect square trinomial. We focus on the $x^2 - 2x$ terms. To complete the square, we take half of the coefficient of our x term (-2), square it ((-1)^2 = 1), and add and subtract it within the expression:

$f(x) = x^2 - 2x + 1 - 1 - 3$

The first three terms now form a perfect square trinomial: $(x - 1)^2$. Our function now appears as $f(x) = (x - 1)^2 - 4$. This, at last, is the vertex form we seek! It elegantly showcases the vertex of the parabola at (1, -4), a testament to the power of algebraic manipulation.

The Vertex Form Unveiled: $f(x) = (x - 1)^2 - 4$

Our meticulous journey has led us to the vertex form: $f(x) = (x - 1)^2 - 4$. Comparing this triumphant result with the options presented, we find a perfect match in option (c). Thus, option (c) stands as the correct representation of our quadratic function in vertex form.

The vertex form, $f(x) = (x - 1)^2 - 4$, not only reveals the vertex at (1, -4) but also the axis of symmetry, the vertical line x = 1, which divides the parabola into two symmetrical halves. The 'a' value of 1 indicates that the parabola opens upwards, and the vertex is its minimum point. This form paints a vivid picture of the parabola's behavior and characteristics.

Conclusion: The Power of Vertex Form

In conclusion, by adeptly utilizing the zeros and a point on the quadratic function, we navigated through factored form, standard form, and ultimately, the coveted vertex form. The journey highlighted the versatility of different forms and the transformative power of completing the square. The vertex form, $f(x) = (x - 1)^2 - 4$, not only answers the specific question but also unlocks a deeper understanding of the parabola's geometry and behavior. This exercise underscores the importance of mastering quadratic functions and their various representations, a cornerstone of mathematical proficiency.

Thus, the equation that faithfully represents the quadratic function with zeros at -1 and 3, gracefully passing through the point (4, 5), in vertex form, is indeed $f(x) = (x - 1)^2 - 4$. This equation encapsulates the essence of the parabola, its vertex, its symmetry, and its trajectory across the coordinate plane.