Question

The graph of the quadratic function \(\mathrm{y}=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\) is a parabola. Find the equation of a parabola passing through the points \((-1,11),(1,3)\), and \((2,5)\), by determining the values of \(\mathrm{a}, \mathrm{b}\), and \(\mathrm{c}\) from the given data.

Short answer

The equation of the parabola passing through the points (-1, 11), (1, 3), and (2, 5) is \(y = 2x^2 - 4x + 5\).

Step-by-step solution

Plug in the given points into the equation

Plugging the point (-1, 11) into the equation, we get: \(11 = a(-1)^2 + b(-1) + c\) Simplifying the equation, we get: \(11 = a - b + c\) Similarly, plugging the point (1, 3) into the equation, we get: \(3 = a(1)^2 + b(1) + c\) Simplifying the equation, we get: \(3 = a + b + c\) Finally, plugging the point (2, 5) into the equation, we get: \(5 = a(2)^2 + b(2) + c\) Simplifying the equation, we get: \(5 = 4a + 2b + c\) Now, we have three linear equations with three variables to solve: 1) \(11 = a - b + c\) 2) \(3 = a + b + c\) 3) \(5 = 4a + 2b + c\)

Solve the system of linear equations

To solve this system, we can eliminate one variable at a time. Let's eliminate the c variable first. We can do this by subtracting equation 1 from equation 2, and equation 1 from equation 3. Subtracting equation 1 from equation 2, we get: \((3 - 11) = (a - a) + (b - (-b)) + (c - c)\) Simplifying, we get: \(-8 = 2b \to b = -4\) Subtracting equation 1 from equation 3, we get: \((5 - 11) = (4a - a) + (2b - (-b)) + (c - c)\) Simplifying and substituting the value of b, we get: \(-6 = 3a + 3( -4) \) Now, solve for a: \(-6 = 3a - 12 \to a = 2\)

Find the value of c

Now that we have the values of a and b, we can find the value of c using any of the three given equations. Let's use the first equation: \(11 = a - b + c\) Substitute the values of a and b: \(11 = 2 - (-4) + c\) Solve for c: \(11 = 6 + c \to c = 5\)

Write the equation of the parabola

Now that we have the values of a, b, and c, we can write the equation of the parabola: \(y = ax^2 + bx + c\) Substitute the values we found: \(y = 2x^2 - 4x + 5\) The equation of the parabola passing through the points (-1, 11), (1, 3), and (2, 5) is \(y = 2x^2 - 4x + 5\).

Key concepts

Parabola

When we talk about quadratic functions, they are often represented graphically by a curve known as a parabola. A parabola is a symmetrical shape that can appear to open upwards or downwards on a graph, depending on its equation. The standard form of a quadratic function is \( y = ax^2 + bx + c \). Here, \(a\), \(b\), and \(c\) are constants.

• If \(a > 0\), the parabola opens upwards like a regular "U" shape.
• If \(a < 0\), it opens downwards.

The "U" shaped curve can help us find important features like the vertex, the axis of symmetry, and the points where the graph crosses the x-axis, called the roots or zeros. In this exercise, our task was to use given points to find the quadratic function, essentially describing where this parabola lies on the coordinate plane.

System of Linear Equations

In this problem, to find the values of \(a\), \(b\), and \(c\), we encountered a system of linear equations. This happens when we have more than one equation working together to find common variable values. Here are the equations from the problem:

• \(11 = a - b + c\)
• \(3 = a + b + c\)
• \(5 = 4a + 2b + c\)

Systems of equations can be solved using various methods, but they must be consistent (work together without contradictions) to find solutions. In our situation, solving this system helped determine the specific coefficients \(a\), \(b\), and \(c\) that made the quadratic function match the provided points.

Substitution Method

The substitution method is a common technique for solving systems of equations. The idea is to solve one equation for one variable and then substitute that solution into the other equations.
In this exercise, we started by eliminating variable \(c\) to simplify the equations. By manipulating the equations through subtraction, we isolated variables and found:

• \(b = -4\)
• Then \(a = 2\)

Once we found these values, they were substituted back to find the final missing variable, \(c\). This method is particularly helpful when dealing with straightforward linear relationships, allowing us to simply replace one variable with its determined value and sequentially solve for others.

Points on a Graph

Points on a graph are typically represented as coordinate pairs \((x, y)\). They tell us exactly where to place a dot on the coordinate plane. For a quadratic equation like \(y = ax^2 + bx + c\), any given point \((x,y)\) should satisfy the equation.
In this exercise, we used the coordinates \((-1, 11)\), \((1, 3)\), and \((2, 5)\) as a basis. Plugging these points into the quadratic equation tells us which values of \(a\), \(b\), and \(c\) are correct for our parabola.

• Each point provides one equation.
• Using multiple points ensures we accurately shape the entire curve.

By ensuring the points satisfy the equation, we verify that the derived quadratic formula is indeed correct, describing a parabola that passes through all three given points.