Question

Let \(\mathrm{f}\) be a function. Show that there is a parabola \(\mathrm{y}=\mathrm{Ax}^{2}+\mathrm{Bx}+\mathrm{C}\) that passes through the three points \((-\mathrm{h}, \mathrm{f}(-\mathrm{h})),(0, \mathrm{f}(0))\), and \((\mathrm{h}, \mathrm{f}(\mathrm{h}))\).

Short answer

Question: Show that there is a parabola y = Ax^2 + Bx + C that passes through the three points with coordinates (-h, f(-h)), (0, f(0)), and (h, f(h)). Answer: By substituting the points into the equation y = Ax^2 + Bx + C and solving the system of equations, we found the values of A, B, and C as: A = (hf(h) - hf(-h)) / 2h^3 B = (f(h) - Ah^2 - f(0)) / h C = f(0) This shows that a parabola y = Ax^2 + Bx + C indeed passes through the points (-h, f(-h)), (0, f(0)), and (h, f(h)).

Step-by-step solution

Substitute the points into the parabola equation

Plug in the coordinates of each of the three points into the parabola equation y = Ax^2 + Bx + C. Doing so will give us three equations as follows: 1) f(-h) = A(-h)^2 + B(-h) + C 2) f(0) = A(0)^2 + B(0) + C 3) f(h) = A(h)^2 + B(h) + C

Simplify the equations

We can simplify these three equations, which would look like this: 1) f(-h) = Ah^2 - Bh + C 2) f(0) = C 3) f(h) = Ah^2 + Bh + C

Solve the system of equations

Since we have f(0) = C from equation (2), we can substitute it back into equations (1) and (3): 1) f(-h) = Ah^2 - Bh + f(0) 3) f(h) = Ah^2 + Bh + f(0) Now, we can solve for B in each of the above equations 1) B = (f(-h) - Ah^2 - f(0)) / -h 3) B = (f(h) - Ah^2 - f(0)) / h Since both of above equations give values for B, we can set them equal to each other, and solve for A: (f(-h) - Ah^2 - f(0)) / -h = (f(h) - Ah^2 - f(0)) / h

Solve for A

To solve for A, we can cross multiply and simplify: h(f(-h) - Ah^2 - f(0)) = -h(f(h) - Ah^2 - f(0)) Clearing the denominators, we get: hf(-h) - Ah^3 - hf(0) = -hf(h) + Ah^3 - hf(0) Combine the terms: 2Ah^3 = hf(h) - hf(-h) Now, divide by 2h^3: A = (hf(h) - hf(-h)) / 2h^3

Find B and C using A

Now that we have found A, we can substitute back into the equations we found for B and C: C = f(0) B = (f(h) - Ah^2 - f(0)) / h This shows that there exists a parabola y = Ax^2 + Bx + C that passes through the points (-h, f(-h)), (0, f(0)), and (h, f(h)).

Key concepts

Parabola Graph

When we discuss the parabola graph, we are visualizing quadratic functions—the U-shaped curves that are symmetrical along a vertical line called the axis of symmetry. This geometric figure can be expressed as an equation in the form of \(y = Ax^2 + Bx + C\), where \(A\), \(B\), and \(C\) are constants and determine the parabola’s shape and position.

A parabola can open upwards or downwards, which is determined by the sign of the coefficient \(A\). If \(A\) is positive, the parabola opens upward, and if \(A\) is negative, it opens downward. The point where the parabola changes direction is known as the vertex. By plotting points and drawing the shape that connects them, students can visually understand how the quadratic equation relates to the corresponding graph.

In relation to the exercise, when showing that a parabola passes through three specific points, one can start by plugging in the coordinates of these points into the general formula to derive a system of equations. Afterward, these points can be plotted to confirm that the graph indeed passes through each one, demonstrating the intimate connection between algebraic expressions and their geometric interpretations.

System of Equations

Solving a system of equations means finding the values for the variables that satisfy all equations simultaneously. In the context of parabolas and quadratic functions, these variables are generally the coefficients of the equation \(y = Ax^2 + Bx + C\).

A system is created by substituting known points on the curve into this equation to create separate equations that are linked together by shared variables. In the given exercise, we used three points to form three equations. Then we aimed to find values for \(A\), \(B\), and \(C\) that make all three equations true at once.

To solve such a system of equations, one may use methods like substitution, elimination, or matrix operations. In the exercise, we effectively used substitution by isolating \(C\) in one equation, then using that value to find \(B\) and \(A\) in the other equations. Eventually, by setting the equations for \(B\) equal to each other, we solved for \(A\), which then allowed us to derive the values of all three coefficients, proving the existence of a parabola through the given points.

Quadratic Functions

Quadratic functions represent a major category of polynomial functions, specifically second-degree polynomials, and they naturally describe parabolic shapes in their graphs. A standard quadratic function is written as \(y = Ax^2 + Bx + C\), where \(A\), \(B\), and \(C\) are constants that affect the shape and position of the parabola as mentioned earlier.

Understanding how each coefficient influences the parabola’s graph is key. The \(A\) coefficient impacts the width and direction of the parabola, \(B\) affects the horizontal location of the vertex, and \(C\) determines the vertical location of the vertex.

These functions are essential in various areas of mathematics and science because they model many real-world phenomena, such as projectile motion. By working through the problem of finding a parabola that passes through three specific points, students gain practical insight into how changes in these coefficients alter the function's graph and how quadratic functions can be used to model real situations.