Finding the vertex of a parabola example | Quadratic equations | Algebra I

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In this lesson, we explored how to find the vertex of a quadratic function, which is represented by the equation \( y = ax^2 + bx + c \). We learned two methods for determining the vertex: using the vertex formula \( x = -\frac{b}{2a} \) and completing the square, both of which revealed that the vertex of the example function \( y = 5x^2 – 20x + 15 \) is located at the point \( (2, -5) \). Understanding these techniques is essential for analyzing the characteristics of parabolas in algebra.

Understanding Quadratic Functions: Finding the Vertex of a Parabola

Quadratic equations are an important part of algebra, and they are usually written in the form ( ax^2 + bx + c ). The graph of a quadratic equation is called a parabola, which can open either upwards or downwards. In this article, we’ll learn how to find the vertex of a quadratic function, which is the highest or lowest point on the parabola.

Characteristics of a Parabola

A quadratic function can be written as ( y = ax^2 + bx + c ). The direction in which the parabola opens is determined by the coefficient ( a ):

  • If ( a > 0 ), the parabola opens upwards, and the vertex is the lowest point.
  • If ( a < 0 ), the parabola opens downwards, and the vertex is the highest point.

For example, if we have a quadratic equation where ( a = 5 ), the parabola opens upwards.

Finding the Vertex Using the Formula

To find the x-coordinate of the vertex, we use the formula:

( x = -frac{b}{2a} )

For our example, the coefficients are:

  • ( b = -20 )
  • ( a = 5 )

Plugging these values into the formula gives:

( x = -frac{-20}{2 times 5} = frac{20}{10} = 2 )

Now that we have the x-coordinate, we can find the y-coordinate by substituting ( x = 2 ) back into the original equation:

( y = 5(2^2) – 20(2) + 15 )

( y = 5(4) – 40 + 15 )

( y = 20 – 40 + 15 = -5 )

So, the vertex of the parabola is at the point ( (2, -5) ).

Completing the Square to Find the Vertex

Besides using the vertex formula, we can also find the vertex by completing the square, which helps us understand the function better.

Starting with the quadratic equation:

( y = 5x^2 – 20x + 15 )

We factor out the coefficient of ( x^2 ) from the first two terms:

( y = 5(x^2 – 4x) + 15 )

Next, we complete the square inside the parentheses. We take half of the coefficient of ( x ) (which is -4), square it, and add it inside the parentheses:

(left(-frac{4}{2}right)^2 = 4)

To keep the equation balanced, we adjust it by subtracting 20 (since we added 4 inside the parentheses, multiplied by 5):

( y = 5(x^2 – 4x + 4) + 15 – 20 )

( y = 5(x – 2)^2 – 5 )

Now, the equation is in vertex form, ( y = a(x – h)^2 + k ), where ( (h, k) ) is the vertex. From this form, we see that the vertex is at ( (2, -5) ).

Conclusion

In conclusion, we learned two ways to find the vertex of a quadratic function: using the vertex formula and completing the square. Both methods show that the vertex of the parabola given by the equation ( y = 5x^2 – 20x + 15 ) is at the point ( (2, -5) ). Understanding these methods is important for analyzing quadratic functions in math.

  1. Reflect on your understanding of quadratic functions before reading the article. How has your perspective changed after learning about finding the vertex of a parabola?
  2. What aspects of the vertex formula did you find most intuitive or challenging, and why?
  3. How does knowing the vertex of a parabola help in understanding the behavior of a quadratic function?
  4. In what ways do you think completing the square provides a deeper understanding of quadratic functions compared to using the vertex formula?
  5. Can you think of any real-world applications where finding the vertex of a parabola might be useful? Share your thoughts.
  6. How might the methods discussed in the article be applied to solve problems in other areas of mathematics or science?
  7. What additional questions do you have about quadratic functions or parabolas after reading the article?
  8. Consider a quadratic function you have encountered before. How would you apply the methods from the article to find its vertex?
  1. Vertex Formula Practice

    Use the vertex formula ( x = -frac{b}{2a} ) to find the vertex of different quadratic equations. Try equations like ( y = 3x^2 + 6x + 2 ) and ( y = -2x^2 + 4x – 1 ). Calculate the x-coordinate and then find the y-coordinate by substituting back into the equation. Share your results with the class.

  2. Graphing Parabolas

    Graph the quadratic equation ( y = 5x^2 – 20x + 15 ) using graph paper or a graphing tool. Identify the vertex and the direction in which the parabola opens. Compare your graph with classmates to see different approaches to graphing.

  3. Completing the Square Challenge

    Work in pairs to complete the square for the quadratic equation ( y = 2x^2 – 8x + 3 ). Rewrite the equation in vertex form and identify the vertex. Discuss the steps with your partner to ensure you both understand the process.

  4. Parabola Art Project

    Create a piece of art using parabolas. Use different quadratic equations to draw parabolas that open upwards and downwards. Label the vertex of each parabola and explain the equation used. Display your artwork in the classroom.

  5. Real-World Quadratics

    Research a real-world application of quadratic functions, such as projectile motion or satellite dishes. Present how the vertex of the parabola is important in that context. Share your findings with the class in a short presentation.

QuadraticA type of polynomial equation of the second degree, generally in the form ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0. – The quadratic equation x² – 4x + 4 = 0 can be solved by factoring.

FunctionA relation between a set of inputs and a set of permissible outputs, typically defined by an equation. – The function f(x) = 2x + 3 represents a linear relationship between x and f(x).

VertexThe highest or lowest point on the graph of a parabola, where the direction changes. – The vertex of the parabola y = x² – 4x + 3 is at the point (2, -1).

ParabolaThe graph of a quadratic function, which is a U-shaped curve that can open upwards or downwards. – The parabola y = -x² + 4x – 3 opens downwards because the coefficient of x² is negative.

CoefficientA numerical or constant factor in front of a variable in an algebraic expression. – In the expression 3x² + 2x + 1, the coefficient of x² is 3.

FormulaA mathematical rule expressed in symbols, often used to calculate a specific value. – The quadratic formula, x = (-b ± √(b² – 4ac)) / (2a), is used to find the roots of a quadratic equation.

X-coordinateThe horizontal value in a pair of coordinates, indicating how far along the point is on the x-axis. – In the point (3, 5), the x-coordinate is 3.

Y-coordinateThe vertical value in a pair of coordinates, indicating how far up or down the point is on the y-axis. – In the point (3, 5), the y-coordinate is 5.

CompletingA method used to solve quadratic equations by transforming them into perfect square trinomials. – By completing the square, the equation x² + 6x + 5 can be rewritten as (x + 3)² – 4.

SquareThe result of multiplying a number by itself, or a term raised to the power of two. – The square of 4 is 16, as 4² = 16.

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