01 76 38 08 47
Kartable logo
HomeBrowseSearchLog in

To enjoy 10 free documents.

Kartable logo
HomeBrowseSearchLog in

To enjoy 10 free documents.

  1. Home
  2. 12th grade
  3. Algebra I
  4. Exercise : Convert between the expanded form and the vertex form of a quadratic function

Convert between the expanded form and the vertex form of a quadratic function Algebra I

Find the vertex form of the following quadratic function:

2x^2-3x-2

The vertex form of a quadratic function is:

y=a\left(x-h\right)^{2}+k

Where \left(h,k\right) are the coordinates of the vertex.

Factor out the coefficient of x^{2}, in our problem 2:

2\left( x^{2}-\dfrac{3}{2}x-1 \right)

Complete a perfect square in the parenthesis:

2\left( x^{2}-2\cdot\dfrac{3}{4}x-1 \right)

2\left( x^{2}-2\cdot\dfrac{3}{4}x+\left( \dfrac{3}{4} \right)^{2}-\left( \dfrac{3}{4} \right)^{2}-1 \right)

Write the perfect square:

2\left[ \left( x-\dfrac{3}{4} \right)^{2}-\dfrac{9}{16}-1 \right]

Simplify the remaining part of the parenthesis:

2\left[ \left( x-\dfrac{3}{4} \right)^{2}-\dfrac{25}{16} \right]

Distribute the 2:

2\left( x-\dfrac{3}{4} \right)^{2}-\dfrac{25}{8}

2x^2-3x-2=2\left( x-\dfrac{3}{4} \right)^{2}-\dfrac{25}{8}

Find the vertex form of the following quadratic function:

5x^2-3x+7

The vertex form of a quadratic function is:

y=a\left(x-h\right)^{2}+k

Where \left(h,k\right) are the coordinates of the vertex.

Factor out the coefficient of x^{2}, in our problem 5:

5\left( x^{2}-\dfrac{3}{5}x+\dfrac{7}{5} \right)

Complete a perfect square in the parenthesis:

5\left( x^{2}-2\cdot\dfrac{3}{10}x+\dfrac{7}{5} \right)

5\left( x^{2}-2\cdot\dfrac{3}{10}x+\left( \dfrac{3}{10} \right)^{2}-\left( \dfrac{3}{10} \right)^{2}+\dfrac{7}{5} \right)

Write the perfect square:

5\left[ \left( x-\dfrac{3}{10} \right)^{2}-\dfrac{9}{100}+\dfrac{7}{5} \right]

Simplify the remaining part of the parenthesis:

5\left[ \left( x-\dfrac{3}{10} \right)^{2}+\dfrac{131}{100} \right]

Distribute the 5:

5\left( x-\dfrac{3}{10} \right)^{2}+\dfrac{131}{20}

5x^{2}-3x+7=5\left( x-\dfrac{3}{10} \right)^{2}+\dfrac{131}{20}

Find the vertex form of the following quadratic function:

3x^2+5x+1

The vertex form of a quadratic function is:

y=a\left(x-h\right)^{2}+k

Where \left(h,k\right) are the coordinates of the vertex.

Factor out the coefficient of x^{2}, in our problem 3:

3\left( x^{2}+\dfrac{5}{3}x+\dfrac{1}{3} \right)

Complete a perfect square in the parenthesis:

3\left( x^{2}+2\cdot\dfrac{5}{6}x+\dfrac{1}{3} \right)

3\left( x^{2}+2\cdot\dfrac{5}{6}x+\left( \dfrac{5}{6} \right)^{2}-\left( \dfrac{5}{6} \right)^{2}+\dfrac{1}{3} \right)

Write the perfect square:

3\left[ \left( x+\dfrac{5}{6} \right)^{2}-\dfrac{25}{36}+\dfrac{1}{3} \right]

Simplify the remaining part of the parenthesis:

3\left[ \left( x+\dfrac{5}{6} \right)^{2}-\dfrac{13}{36} \right]

Distribute the 3:

3\left( x+\dfrac{5}{6} \right)^{2}-\dfrac{13}{12}

3x^2+5x+1=3\left( x+\dfrac{5}{6} \right)^{2}-\dfrac{13}{12}

Find the vertex form of the following quadratic function:

2x^2+16x-15

The vertex form of a quadratic function is:

y=a\left(x-h\right)^{2}+k

Where \left(h,k\right) are the coordinates of the vertex.

Factor out the coefficient of x^{2}, in our problem 2:

2\left( x^{2}+8x-\dfrac{15}{2} \right)

Complete a perfect square in the parenthesis:

2\left( x^{2}+2\cdot4x-\dfrac{15}{2} \right)

2\left( x^{2}+2\cdot4x+\left( 4 \right)^{2}-\left( 4 \right)^{2}-\dfrac{15}{2} \right)

Write the perfect square:

2\left[ \left( x+4 \right)^{2}-16-\dfrac{15}{2} \right]

Simplify the remaining part of the parenthesis:

2\left[ \left( x+4 \right)^{2}-\dfrac{47}{2} \right]

Distribute the 2:

2\left( x+4 \right)^{2}-47

2x^{2}+16x-15=2\left( x+4 \right)^{2}-47

Find the standard form of the following quadratic function:

2\left(x-5\right)^{2}+4

In order to find the standard form of a quadratic function, simplify the given expression:

2\left(x-5\right)^{2}+4=2\left(x-5\right)\left(x-5\right)+4

2\left(x-5\right)^{2}+4=2\left(x^{2}-5x-5x+25\right)+4

2\left(x-5\right)^{2}+4=2\left(x^{2}-10x+25\right)+4

2\left(x-5\right)^{2}+4=2x^{2}-20x+50+4

2\left(x-5\right)^{2}+4=2x^{2}-20x+54

Find the standard form of the following quadratic function:

9\left( x-\dfrac{1}{3} \right)^{2}-7

In order to find the standard form of a quadratic function, simplify the given expression:

9\left( x-\dfrac{1}{3} \right)^{2}-7=9\left( x-\dfrac{1}{3} \right)\left( x-\dfrac{1}{3} \right)-7

9\left( x-\dfrac{1}{3} \right)^{2}-7=9\left( x^{2}-\dfrac{1}{3}x-\dfrac{1}{3}x+\dfrac{1}{9} \right)-7

9\left( x-\dfrac{1}{3} \right)^{2}-7=9\left( x^{2}-\dfrac{2}{3}x+\dfrac{1}{9} \right)-7

9\left( x-\dfrac{1}{3} \right)^{2}-7=9x^{2}-6x+1-7

9\left( x-\dfrac{1}{3} \right)^{2}-7=9x^{2}-6x-6

Find the standard form of the following quadratic function:

3\left( x+\dfrac{1}{6} \right)^{2}-\dfrac{133}{12}

In order to find the standard form of a quadratic function, simplify the given expression:

3\left( x+\dfrac{1}{6} \right)^{2}-\dfrac{133}{12}=3\left( x+\dfrac{1}{6} \right)\left( x+\dfrac{1}{6} \right)-\dfrac{133}{12}

3\left( x+\dfrac{1}{6} \right)^{2}-\dfrac{133}{12}=3\left( x^{2}+\dfrac{1}{6}x+\dfrac{1}{6}x+\dfrac{1}{36} \right)-\dfrac{133}{12}

3\left( x+\dfrac{1}{6} \right)^{2}-\dfrac{133}{12}=3\left( x^{2}+\dfrac{1}{3}x+\dfrac{1}{36} \right)-\dfrac{133}{12}

3\left( x+\dfrac{1}{6} \right)^{2}-\dfrac{133}{12}=3x^{2}+x+\dfrac{1}{12}-\dfrac{133}{12}

3\left( x+\dfrac{1}{6} \right)^{2}-\dfrac{133}{12}=3x^{2}+x-11

The editorial charter guarantees the compliance of the content with the official National Education curricula. Learn more

The courses and exercises are written by the Kartable editorial team, made up of teachers certified and accredited. Learn more

See also
  • Course : Quadratic functions
  • Exercise : Match curves and quadratic functions
  • Exercise : Calculate the average rate of change of a quadratic function
  • Exercise : Solve quadratic equations with calculations
  • Exercise : Solve quadratic inequalities with calculations
  • support@kartable.com
  • Legal notice

© Kartable 2026