Problem 98 - Precalculus

Question

Find the values of \(a, b,\) and \(c\) such that the graph of the quadratic function \(y=a x^{2}+b x+c\) passes through the points \((1,5),(-2,-10),\) and (0,4)

Step-by-Step Solution

Answer

The values are \(a = -2\), \(b = 3\), and \(c = 4\).

Step 1: Substitute the Points into the Quadratic Equation

We are given three points: \((1,5)\), \((-2,-10)\), and \((0,4)\). Substitute each point into the quadratic equation \(y = ax^2 + bx + c\) to get three equations.For the point \((1,5)\): \[5 = a(1)^2 + b(1) + c\] Simplifies to:\[a + b + c = 5\]For the point \((-2,-10)\): \[-10 = a(-2)^2 + b(-2) + c\] Simplifies to:\[4a - 2b + c = -10\]For the point \((0,4)\):\[4 = a(0)^2 + b(0) + c\] Simplifies to:\[c = 4\]

Step 2: Substitute c and Simplify Equations

From the third equation in Step 1, we know that \(c = 4\). Substitute \(c = 4\) into the other two equations:First equation becomes: \[a + b + 4 = 5\] Simplifies to: \[a + b = 1\]Second equation becomes: \[4a - 2b + 4 = -10\] Simplifies to: \[4a - 2b = -14\]

Step 3: Solve the System of Equations for a and b

We have a system of two equations: 1. \(a + b = 1\) 2. \(4a - 2b = -14\)First, solve equation (1) for \(b\): \[b = 1 - a\]Substitute \(b = 1 - a\) into equation (2): \[4a - 2(1 - a) = -14\]Simplifies to: \[4a - 2 + 2a = -14\] \[6a - 2 = -14\]Add 2 to both sides:\[6a = -12\]Divide by 6:\[a = -2\]

Step 4: Determine the Value of b

With \(a = -2\), substitute back to find \(b\):\[b = 1 - a\]\[b = 1 - (-2)\]\[b = 1 + 2\]\[b = 3\]

Step 5: Conclude the Values for a, b, and c

We have found all the values:\(a = -2\), \(b = 3\), and \(c = 4\). These values satisfy the given condition that the quadratic function passes through the three specified points.

Key Concepts

Understanding Quadratic Equations

Quadratic equations are a central topic in algebra. They represent polynomial equations of the form \(y = ax^2 + bx + c\). Here, \(a\), \(b\), and \(c\) are constants with \(a eq 0\). This constraint ensures the "quadratic" nature of the function, which means it involves an \(x^2\) term.

• **Why \(a eq 0\)?** If \(a = 0\), the equation becomes linear, taking the form \(y = bx + c\).
• **Roots of the quadratic equation:** These are the solutions for \(x\) when \(y = 0\).
• **Standard way to solve:** Techniques to solve include factoring, completing the square, or using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

In many problems like the one given, you often need to find the specific coefficients \(a\), \(b\), and \(c\) by using known points the graph passes through. For such problems, substituting these points into the equation helps form a system of equations that can be solved to identify the values of \(a\), \(b\), and \(c\).
Understanding and solving quadratic equations is fundamental, as they appear frequently in various mathematical settings, from physics to engineering.

Solving Systems of Equations

A system of equations consists of multiple equations that share the same set of variables. The goal is to find the values of these variables that satisfy all equations simultaneously. For the given exercise, substituting the known points into the quadratic equation formed a **system of equations**.

• **What it means:** In our problem, each point provides an equation, all of which must hold true concurrently. Hence, determining \(a\), \(b\), and \(c\) required solving this system.
• **Typical solving strategies:** For linear systems, common methods include substitution, elimination, or using matrices. In this case, substitution was applied:
  • First, substitute values to express one variable in terms of another.
  • Use this expression to replace the variable in the other equations.

The solution for \(a\) and \(b\) involved expressing one in terms of the other and substituting back into another equation until all variables have explicit values. Success with systems of equations sets a solid foundation for linear algebra concepts encountered later in advanced math.

Graphing Quadratics

Graphing quadratics is a visual way to understand the behavior of quadratic functions. The graph of a quadratic equation \(y = ax^2 + bx + c\) forms a parabola.

• **Direction of the parabola:** If \(a > 0\), the parabola opens upwards. If \(a < 0\), like in our problem, it opens downwards.
• **Vertex and axis of symmetry:** The vertex is the parabola's highest or lowest point, found at \(x = \frac{-b}{2a}\). This line, \(x = \frac{-b}{2a}\), is also the axis of symmetry, where the parabola mirrors itself.
• **Y-intercept:** Directly given by \(c\), where the graph intersects the y-axis. In the problem, \(c = 4\), showing the graph crosses the y-axis at \(y = 4\).

Graphing helps illustrate how variations in \(a\), \(b\), and \(c\) can affect the parabola's shape and position. Matching a parabola to points involves adjusting these parameters to ensure that the graph represents the specific quadratic relationship accurately. Being able to interpret and sketch graphs of quadratics is an invaluable skill in math and science applications.