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Chapter 10: Problem 36

Find an equation for the conic that satisfies the given conditions. $$ \begin{array}{l}{\text { Parabola, vertical axis, }} \\ {\text { passing through }(0,4),(1,3), \text { and }(-2,-6)}\end{array} $$

Short Answer

Expert verified
The equation is \(y = -2x^2 + x + 4\).

Step by step solution

01

Recognize the Parabola Equation

Since the parabola has a vertical axis, its general equation is of the form \(y = ax^2 + bx + c\). We need to find the constants \(a\), \(b\), and \(c\) using the given points.
02

Substitute the Points into the Equation

Use the given points to create a system of equations. For the point \((0, 4)\), substitute into \(y = ax^2 + bx + c\) to get the equation \(4 = c\). For the point \((1, 3)\), substitute to get \(3 = a(1)^2 + b(1) + c\), leading to \(3 = a + b + c\). For the point \((-2, -6)\), substitute to get \(-6 = a(-2)^2 + b(-2) + c\), leading to \(-6 = 4a - 2b + c\).
03

Solve the Equation for c

From the first equation \(4 = c\), we immediately know \(c = 4\). Substitute this value into the other two equations.
04

Substitute c into the Remaining Equations

Substitute \(c = 4\) into \(3 = a + b + c\) to get \(3 = a + b + 4\), which simplifies to \(a + b = -1\). Also, substitute \(c = 4\) into \(-6 = 4a - 2b + c\) to get \(-6 = 4a - 2b + 4\), leading to \(4a - 2b = -10\).
05

Solve the System of Equations for a and b

We now have two equations: \(a + b = -1\) and \(4a - 2b = -10\). Solve using substitution or elimination. For elimination, multiply the first equation by 2 to get \(2a + 2b = -2\). Add it to \(4a - 2b = -10\) to eliminate \(b\): \(6a = -12\), so \(a = -2\). Substitute \(a = -2\) back into \(a + b = -1\) to get \(-2 + b = -1\), so \(b = 1\).
06

Write the Final Equation of the Parabola

Having found \(a = -2\), \(b = 1\), and \(c = 4\), substitute back into the general equation to get the equation of the parabola: \(y = -2x^2 + x + 4\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertical Axis
A parabola with a vertical axis signifies that it opens up or down as opposed to sideways.
This orientation affects the general form of its equation.
  • The standard equation for a vertically oriented parabola is \( y = ax^2 + bx + c \).
  • The axis of symmetry for such a parabola is a vertical line \(x = -\frac{b}{2a} \).
  • The vertex of the parabola lies on this axis.
When solving problems involving parabolas, knowing the orientation helps us decide which form of the equation to use.
For a parabola with a vertical axis, we focus on solving for coefficients \(a\), \(b\), and \(c\) in the given equation.
System of Equations
In this exercise, we find the parabola's equation by creating a system of equations using the given points. Each point is a pair of x, y values that satisfy the parabola's equation.
  • By substituting the x and y coordinates of each point into the equation \( y = ax^2 + bx + c \), we form a separate equation for each.
  • In our example, substituting the points \((0, 4)\), \((1, 3)\), and \((-2, -6)\) yields three equations.
  • We then solve this system to find the values of \(a\), \(b\), and \(c\).
Solving a system of equations often requires methods like substitution or elimination, essential for finding the best-fit curve that passes through given points.
General Equation of Parabola
The general equation of a parabola with a vertical axis of symmetry is \( y = ax^2 + bx + c \). This form is versatile and allows for easy computation of parabola characteristics.
  • \(a\), \(b\), and \(c\) determine the parabola's shape and position:
  • The value of \(a\) dictates the direction of the opening (upwards if \(a > 0\), downwards if \(a < 0\)).
  • The coefficient \(b\) affects the direction of the vertex from the y-axis.
  • \(c\) provides the y-intercept, indicating the vertical shift of the parabola.
Knowing how to manipulate these constants helps in graphing and understanding shifts, stretches, and orientations of parabolas.
Solving Quadratic Equations
When solving a quadratic equation like \( y = ax^2 + bx + c \), we aim to find the values of \(x\) that satisfy the equation. This process involves several algebraic methods.
  • For simple equations, we attempt direct factoring if it's possible.
  • If factoring is challenging, the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) provides solutions for any quadratic equation.
  • An alternative is completing the square, which involves rewriting the quadratic in a perfect square form to solve for \(x\).
Understanding these methods will help solve any quadratic equation, leading to finding the x-values where the parabola intersects the x-axis or to solve systems of equations involving quadratic expressions.

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Most popular questions from this chapter

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