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Writing the equation of a quadratic function from the graph

Write an equation for the quadratic function g in [link] as a transformation of f ( x ) = x 2 , and then expand the formula, and simplify terms to write the equation in general form.

Graph of a parabola with its vertex at (-2, -3).

We can see the graph of g is the graph of f ( x ) = x 2 shifted to the left 2 and down 3, giving a formula in the form g ( x ) = a ( x + 2 ) 2 3.

Substituting the coordinates of a point on the curve, such as ( 0 , −1 ) , we can solve for the stretch factor.

1 = a ( 0 + 2 ) 2 3     2 = 4 a     a = 1 2

In standard form, the algebraic model for this graph is ( g ) x = 1 2 ( x + 2 ) 2 3.

To write this in general polynomial form, we can expand the formula and simplify terms.

g ( x ) = 1 2 ( x + 2 ) 2 3         = 1 2 ( x + 2 ) ( x + 2 ) 3         = 1 2 ( x 2 + 4 x + 4 ) 3         = 1 2 x 2 + 2 x + 2 3         = 1 2 x 2 + 2 x 1

Notice that the horizontal and vertical shifts of the basic graph of the quadratic function determine the location of the vertex of the parabola; the vertex is unaffected by stretches and compressions.

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A coordinate grid has been superimposed over the quadratic path of a basketball in [link] . Find an equation for the path of the ball. Does the shooter make the basket?

Stop motioned picture of a boy throwing a basketball into a hoop to show the parabolic curve it makes.
(credit: modification of work by Dan Meyer)

The path passes through the origin and has vertex at ( 4 ,   7 ) , so ( h ) x = 7 16 ( x + 4 ) 2 + 7. To make the shot, h ( 7.5 ) would need to be about 4 but h ( 7.5 ) 1.64 ; he doesn’t make it.

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Given a quadratic function in general form, find the vertex of the parabola.

  1. Identify a ,   b ,   and   c .
  2. Find h , the x -coordinate of the vertex, by substituting a and b into h = b 2 a .
  3. Find k , the y -coordinate of the vertex, by evaluating k = f ( h ) = f ( b 2 a ) .

Finding the vertex of a quadratic function

Find the vertex of the quadratic function f ( x ) = 2 x 2 6 x + 7. Rewrite the quadratic in standard form (vertex form).

The horizontal coordinate of the vertex will be at

h = b 2 a    = 6 2 ( 2 )    = 6 4    = 3 2

The vertical coordinate of the vertex will be at

k = f ( h )    = f ( 3 2 )    = 2 ( 3 2 ) 2 6 ( 3 2 ) + 7    = 5 2

Rewriting into standard form, the stretch factor will be the same as the a in the original quadratic.

f ( x ) = a x 2 + b x + c f ( x ) = 2 x 2 6 x + 7

Using the vertex to determine the shifts,

f ( x ) = 2 ( x 3 2 ) 2 + 5 2
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Given the equation g ( x ) = 13 + x 2 6 x , write the equation in general form and then in standard form.

g ( x ) = x 2 6 x + 13 in general form; g ( x ) = ( x 3 ) 2 + 4 in standard form

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Finding the domain and range of a quadratic function

Any number can be the input value of a quadratic function. Therefore, the domain of any quadratic function is all real numbers. Because parabolas have a maximum or a minimum point, the range is restricted. Since the vertex of a parabola will be either a maximum or a minimum, the range will consist of all y -values greater than or equal to the y -coordinate at the turning point or less than or equal to the y -coordinate at the turning point, depending on whether the parabola opens up or down.

Domain and range of a quadratic function

The domain of any quadratic function is all real numbers.

The range of a quadratic function written in general form f ( x ) = a x 2 + b x + c with a positive a value is f ( x ) f ( b 2 a ) , or [ f ( b 2 a ) , ) ; the range of a quadratic function written in general form with a negative a value is f ( x ) f ( b 2 a ) , or ( , f ( b 2 a ) ] .

The range of a quadratic function written in standard form f ( x ) = a ( x h ) 2 + k with a positive a value is f ( x ) k ; the range of a quadratic function written in standard form with a negative a value is f ( x ) k .

Practice Key Terms 6

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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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