/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 Find the equation of the parabol... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the equation of the parabola with vertical axis that passes through the data points shown or specified. Check your answer. $$(2,9),(-2,1),(-3,4)$$

Short Answer

Expert verified
The equation is \( y = x^2 + 2x + 1 \).

Step by step solution

01

Understand the Parabola Equation

A parabola with a vertical axis has the general equation \( y = ax^2 + bx + c \). Our goal is to find the coefficients \( a \), \( b \), and \( c \).
02

Set Up the System of Equations

Substitute each point into the equation \( y = ax^2 + bx + c \):For \((2,9)\), we have: \( 9 = 4a + 2b + c \).For \((-2,1)\), we have: \( 1 = 4a - 2b + c \).For \((-3,4)\), we have: \( 4 = 9a - 3b + c \).
03

Solve the System of Equations

We have three equations:1. \( 4a + 2b + c = 9 \)2. \( 4a - 2b + c = 1 \)3. \( 9a - 3b + c = 4 \)Subtract the second equation from the first:\[ 0a + 4b = 8 \Rightarrow b = 2 \]Substitute \( b = 2 \) into the first equation:\[ 4a + 2(2) + c = 9 \Rightarrow 4a + c = 5 \quad (4)\]Substitute \( b = 2 \) into the third equation:\[ 9a - 3(2) + c = 4 \Rightarrow 9a - 6 + c = 4 \Rightarrow 9a + c = 10 \quad (5)\]Subtract equation (4) from equation (5):\[ 5a = 5 \Rightarrow a = 1 \]Substitute \( a = 1 \) into equation (4):\[ 4(1) + c = 5 \Rightarrow c = 1 \]
04

Write the Equation of the Parabola

Now that we have \( a = 1 \), \( b = 2 \), and \( c = 1 \), the equation of the parabola is:\[ y = x^2 + 2x + 1 \]
05

Verify the Solution

Substitute the points into the equation \( y = x^2 + 2x + 1 \):For \((2,9)\): \[ y = (2)^2 + 2(2) + 1 = 4 + 4 + 1 = 9 \]For \((-2,1)\): \[ y = (-2)^2 + 2(-2) + 1 = 4 - 4 + 1 = 1 \]For \((-3,4)\): \[ y = (-3)^2 + 2(-3) + 1 = 9 - 6 + 1 = 4 \]All points satisfy the equation.

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Key Concepts

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

Parabolas
Parabolas are a type of curve on a graph that can open upwards or downwards. They are the graphical representation of quadratic equations, which are equations of the form \(y = ax^2 + bx + c\). A parabola has a vertex, which is the highest or lowest point, and a vertical axis of symmetry, meaning it is mirrored on each side as you move away from the vertex. The shape and position of the parabola are determined by the coefficients \(a\), \(b\), and \(c\). When \(a > 0\), the parabola opens upwards; when \(a < 0\), it opens downwards. This characteristic curve is essential in many real-world applications like projectile motion and satellite dishes.
Key points about parabolas include:
  • The vertex is found using the formula \(x = -\frac{b}{2a}\).
  • The parabola is symmetric around its vertical axis.
  • The y-intercept is the constant \(c\).
System of Equations
A system of equations is a set of two or more equations that have common variables. Solving a system means finding the values of the variables that satisfy all equations in the system simultaneously. In the context of finding the equation of a parabola, we set up a system of equations by inserting the given points into the general form of a quadratic equation \(y = ax^2 + bx + c\).
For example, using our given points \((2,9)\), \((-2,1)\), and \((-3,4)\), we derive three equations:
  • \(4a + 2b + c = 9\)
  • \(4a - 2b + c = 1\)
  • \(9a - 3b + c = 4\)
Each of these equations represents a curve passing through one of the points on the parabola. Solving these simultaneously will help us find the unique values for \(a\), \(b\), and \(c\) that define the parabola.
Substitution Method
The substitution method is a common algebraic technique used to solve systems of equations. In this method, you solve one equation for one of the variables and then substitute that expression into the other equations. This can help reduce the number of variables and make it easier to solve the system.
In our problem, after formulating the system of equations, we first solve \(b\) by subtracting the second equation from the first to get \(b = 2\). Then, we substitute \(b = 2\) into the other equations to further simplify and solve for \(a\) and \(c\).
Using substitution:
  • With \(b = 2\), we found \(4a + c = 5\) and \(9a + c = 10\).
  • By subtracting these, we find \(a = 1\).
  • Finally, substituting \(a = 1\) back into the simpler equation gives us \(c = 1\).
This method is especially useful when dealing with systems of two or more equations.
Verification of Solution
Verification of a solution involves checking whether the obtained values satisfy all the original equations in the system. This is important to ensure that the solution found is correct and consistent.
In our case, the parabola equation we found was \(y = x^2 + 2x + 1\). To verify this, we substitute each of the original points into this equation and check if the equation holds true:
  • For the point \((2,9)\), substitute to check if \(9 = 2^2 + 2(2) + 1\).
  • For \((-2,1)\), substitute and verify \(1 = (-2)^2 + 2(-2) + 1\).
  • For \((-3,4)\), substitute and verify \(4 = (-3)^2 + 2(-3) + 1\).
All these verifications show that the parabola equation correctly passes through all the given points, confirming the accuracy of our solution.

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Most popular questions from this chapter

In certain parts of the Rocky Mountains, deer are the main food source for mountain lions. When the deer population \(d\) is large, the mountain lions ( \(m\) ) thrive. However, a large mountain lion population drives down the size of the deer population. Suppose the fluctuations of the two populations from year to year can be modeled with the matrix equation $$\left[\begin{array}{c} m_{n+1} \\ d_{n+1} \end{array}\right]=\left[\begin{array}{cc} 0.51 & 0.4 \\ -0.05 & 1.05 \end{array}\right]\left[\begin{array}{l} m_{n} \\ d_{n} \end{array}\right]$$ The numbers in the column matrices give the numbers of animals in the two populations after \(n\) years and \(n+1\) years, where the number of deer is measured in hundreds. (a) Give the equation for \(d_{n+1}\) obtained from the second row of the square matrix. Use this equation to determine the rate the deer population will grow from year to year if there are no mountain lions. (b) Suppose we start with a mountain lion population of 2000 and a deer population of \(500,000\) (that is, 5000 hundred deer). How large would each population be after 1 year? 2 years? (c) Consider part (b), but change the initial mountain lion population to \(4000 .\) Show that the populations would both grow at a steady annual rate of \(1 \$ 6\).

Solve each system graphically. Check your solutions. Do not use a calculator. $$\begin{aligned}&x^{2}+y^{2}=5\\\&x+y=3\end{aligned}$$

For what value(s) of \(k\) will the following system of linear equations have no solution? infinitely many solutions? $$\begin{array}{r}x-2 y=3 \\\\-2 x+4 y=k\end{array}$$

As the price of a product increases, businesses usually increase the quantity manufactured. However, as the price increases, consumer demand-or the quantity of the product purchased by consumers-usually decreases. The price we see in the market place occurs when the quantity supplied and the quantity demanded are equal. This price is called the equilibrium price and this demand is called the equilibrium demand. (Refer to Exercise 92 .) Suppose that supply is related to price by \(p=\frac{1}{10} q\) and that demand is related to price by \(p=15-\frac{2}{3} q,\) where \(p\) is price in dollars and \(q\) is the quantity supplied in units. (a) Determine the price at which 15 units would be supplied. Determine the price at which 15 units would be demanded. (b) Determine the equilibrium price at which the quantity supplied and quantity demanded are equal. What is the demand at this price?

Solve each system. Write solutions in terms of \(z\) if necessary. $$\begin{aligned} 3 x+6 y-3 z &=12 \\ -x-2 y+z &=16 \\ x+y-2 z &=20 \end{aligned}$$

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