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Find the equation of the parabola \(y=a x^{2}+b x+c\) that passes through the points \((2,3),(-1,0),\) and \((-2,2)\) Find the equation of the parabola \(y=a x^{2}+b x+c\) that passes through the points \((-2,4),(2,2),\) and \((4,9)\)

Short Answer

Expert verified
The equation of the parabola is \ y = \frac{2}{3} x^2 - \frac{1}{2} x + \frac{1}{3} \.

Step by step solution

01

Set Up the System of Equations

Substitute each given point into the equation of the parabola \(y = ax^2 + bx + c\) to create a system of equations. The points are \((-2, 4)\), \((2, 2)\), and \((4, 9)\). \( \begin{cases} 4 = 4a - 2b + c \ 2 = 4a + 2b + c \ 9 = 16a + 4b + c \ \ \end{cases} \)
02

Subtract Equations to Eliminate c

Subtract the first equation from the second and the second from the third to eliminate \(c\) and simplify the system: \( (2 = 4a + 2b + c) - (4 = 4a - 2b + c) \ \ \Rightarrow -2 = 4b \ \ (9 = 16a + 4b + c) - (2 = 4a + 2b + c) \ \ \Rightarrow 7 = 12a + 2b\)
03

Solve for b

Solve for \(b\) using the simplified system: \ -2 = 4b \ \ \Rightarrow b = -\frac{1}{2}\
04

Substitute b and Solve for a

Substitute \(b = -\frac{1}{2}\) into the second simplified equation to solve for \(a\): \ 7 = 12a + 2\left(-\frac{1}{2}\right)\ \ 7 = 12a - 1 \ \ \Rightarrow a = \frac{2}{3}\
05

Substitute a and b to Solve for c

Substitute \(a = \frac{2}{3}\) and \(b = -\frac{1}{2}\) into one of the original equations to solve for \(c\): \ 4 = 4\left( \frac{2}{3} \right) - 2\left(-\frac{1}{2}\right) + c \ \ \Rightarrow 4 = \frac{8}{3} + 1 + c \ \Rightarrow 4 = \frac{11}{3} + c \ \ \Rightarrow c = 4 - \frac{11}{3} = \frac{1}{3}\
06

Write the Final Equation

With \(a = \frac{2}{3}\), \(b = -\frac{1}{2}\), and \(c = \frac{1}{3}\) the final equation of the parabola is: \ y = \frac{2}{3} x^2 - \frac{1}{2} x + \frac{1}{3} \

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Key Concepts

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

parabolas
A parabola is a symmetrical, curved shape that can open upwards or downwards on a graph. It is represented mathematically by a quadratic equation: y = ax^2 + bx + c.In this equation, 'a', 'b', and 'c' are constants that determine the shape and position of the parabola.

The coefficient 'a' affects the width and direction of the parabola:
  • If 'a' is positive, the parabola opens upwards.
  • If 'a' is negative, the parabola opens downwards.
  • The larger the absolute value of 'a', the narrower the parabola.

The vertex is the highest or lowest point on a parabola, and it represents the maximum or minimum value of the quadratic function. The axis of symmetry of the parabola is a vertical line that passes through the vertex, splitting the parabola into two identical halves.

When graphing a parabola, it's useful to find several important points such as the vertex, the y-intercept (where the parabola crosses the y-axis), and the x-intercepts (where the parabola crosses the x-axis). These points help in accurately plotting the curve. By using the given points, we can establish a system of equations to find the values of 'a,' 'b,' and 'c.'
system of equations
A system of equations is a set of two or more equations with the same variables. Solving a system of equations means finding the values of the variables that satisfy all the equations in the system. In the context of finding the equation of a parabola, we can use the given points to create a system of linear equations.

Let's take the example given in the solution: we have points y = ax^2 + bx + c y = 4 at x = -2 y = 2 at x = 2 y = 9 at x = 4

By substituting these points into the quadratic equation, we get three equations:
  • 4 = 4a - 2b + c
  • 2 = 4a + 2b + c
  • 9 = 16a + 4b + c

We can solve these equations simultaneously to find the values of 'a', 'b', and 'c'. Efficiently solving a system usually involves algebraic methods like substitution, elimination, or matrix operations to reduce the system to simpler forms where solutions can be easily obtained.
substitution method
The substitution method is one of the techniques used to solve a system of equations. This method involves solving one equation for one of the variables and then substituting that expression into the other equations. This helps to reduce the number of variables and simplifies the equations, making them easier to solve.

In our parabola example, after setting up the system of equations:
y = 4a - 2b + cy = 4 at x = -2y = 2 at x = 2y = 9 at x = 4We can subtract the first equation from the second and the second from the third to eliminate 'c':
  • (2 = 4a + 2b + c) - (4 = 4a - 2b + c) → -2 = 4b
  • (9 = 16a + 4b + c) - (2 = 4a + 2b + c) → 7 = 12a + 2b
From these simplified equations, we first solve for 'b':
-2 = 4b → b = -1/2
Then substitute 'b' into the second simplified equation:
7 = 12a - 1 → a = 2/3
Lastly, substitute 'a' and 'b' back into one of the original equations to find 'c':
4 = 4(2/3) - 2(-1/2) + c → c = 1/3
The final quadratic equation is y = (2/3)x^2 - (1/2)x + 1/3.
Using substitution simplifies the solving process considerably, especially when dealing with multiple variables.

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