Function With Minimum At (4, -3): Find The Graph

Hey there, math enthusiasts! Today, we're diving into a fun problem: identifying which quadratic function has its minimum point (also known as the vertex) at the coordinates (4, -3). This involves understanding the relationship between a quadratic function's equation and the location of its vertex. We'll break down the options, explore the key concepts, and arrive at the solution together. So, let's get started!

Understanding Quadratic Functions and Their Minimums

Before we jump into the options, let's quickly review what quadratic functions are and how their minimums (or maximums) are determined. A quadratic function is a polynomial function of degree 2, generally written in the form:

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

where a, b, and c are constants, and a is not equal to 0. The graph of a quadratic function is a parabola, a U-shaped curve. The parabola opens upwards if a > 0, meaning it has a minimum point. Conversely, it opens downwards if a < 0, indicating a maximum point.

The vertex of the parabola is this minimum (or maximum) point. The x-coordinate of the vertex can be found using the formula:

x = -b / 2a

Once we have the x-coordinate, we can substitute it back into the function to find the corresponding y-coordinate, giving us the vertex (x, y).

Key Concepts Recap:

• Quadratic Function: f(x) = ax^2 + bx + c
• Parabola: The U-shaped graph of a quadratic function.
• Vertex: The minimum or maximum point of the parabola.
• Vertex x-coordinate: x = -b / 2a

With these concepts in mind, we can now analyze the given options.

Analyzing the Options

We are given four options, each representing a quadratic function. Our goal is to find the one whose graph has a minimum located at (4, -3). This means we need to check which function has a vertex at this point. Let's look at each option step-by-step:

A. f(x) = -1/2 x^2 + 4x - 11

In this function, a = -1/2, b = 4, and c = -11. Since a is negative, the parabola opens downwards, meaning it has a maximum, not a minimum. Therefore, this option can be eliminated right away.

B. f(x) = -2x^2 + 16x - 35

Here, a = -2, b = 16, and c = -35. Again, a is negative, so the parabola opens downwards and has a maximum, not a minimum. This option is also incorrect.

C. f(x) = 1/2 x^2 - 4x + 5

In this case, a = 1/2, b = -4, and c = 5. Since a is positive, the parabola opens upwards and has a minimum. Let's find the x-coordinate of the vertex:

x = -b / 2a = -(-4) / (2 * 1/2) = 4 / 1 = 4

The x-coordinate matches the given minimum point (4, -3). Now, let's find the y-coordinate by substituting x = 4 into the function:

f(4) = 1/2 (4)^2 - 4(4) + 5 = 1/2 (16) - 16 + 5 = 8 - 16 + 5 = -3

The y-coordinate is -3, which matches the given minimum point. So, this option seems to be correct, but let's check the last option just to be sure.

D. f(x) = 2x^2 - 16x + 35

For this function, a = 2, b = -16, and c = 35. Since a is positive, the parabola opens upwards and has a minimum. Let's find the x-coordinate of the vertex:

x = -b / 2a = -(-16) / (2 * 2) = 16 / 4 = 4

The x-coordinate matches our target. Now, let's find the y-coordinate:

f(4) = 2(4)^2 - 16(4) + 35 = 2(16) - 64 + 35 = 32 - 64 + 35 = 3

The y-coordinate is 3, which does not match the given minimum point of -3. Therefore, this option is incorrect.

Recap of Analysis:

• Options A and B were eliminated because they have maximums, not minimums.
• Option C had a minimum at (4, -3), matching the given point.
• Option D had the correct x-coordinate for the minimum but the wrong y-coordinate.

The Solution

After carefully analyzing each option, we can confidently conclude that the function with a minimum located at (4, -3) is:

C. f(x) = 1/2 x^2 - 4x + 5

Why This Works:

This function satisfies the conditions because:

• It's a quadratic function with a positive leading coefficient (a = 1/2), ensuring it opens upwards and has a minimum.
• The vertex, calculated using x = -b / 2a and then substituting into the function, matches the given point (4, -3).

Final Thoughts

Understanding how the coefficients of a quadratic function relate to its graph, particularly the vertex, is a crucial skill in algebra. By using the vertex formula and carefully evaluating the function, we were able to pinpoint the correct answer from the given options. Remember, practice makes perfect! The more you work with quadratic functions, the more intuitive these concepts will become.

If you want to explore more about quadratic functions and their graphs, check out this helpful resource: Khan Academy on Quadratic Functions.