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5.1 Quadratic functions  (Page 3/15)

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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. First, find the horizontal coordinate of the vertex. Then find the vertical coordinate of the vertex. Substitute the values into standard form, using the “a” from the general form.

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

The standard form of a quadratic function prior to writing the function then becomes the following:

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 unless the context of the function presents some restrictions.

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 .

Given a quadratic function, find the domain and range.

  1. Identify the domain of any quadratic function as all real numbers.
  2. Determine whether   a   is positive or negative. If   a   is positive, the parabola has a minimum. If   a   is negative, the parabola has a maximum.
  3. Determine the maximum or minimum value of the parabola,   k .
  4. If the parabola has a minimum, the range is given by   f ( x ) ≥ k ,   or   [ k , ∞ ) .   If the parabola has a maximum, the range is given by   f ( x ) ≤ k ,   or   ( − ∞ , k ] .

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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