5.1 Quadratic functions (Page 3/15)

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Try It

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 $(h) x = - \frac{7}{16} {(x + 4)}^{2} + 7.$ To make the shot, $h (- 7.5)$ would need to be about 4 but $h (- 7.5) \approx 1.64;$ he doesn’t make it.

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How To

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

Identify $a, b, and c .$
Find $h,$ the x -coordinate of the vertex, by substituting $a$ and $b$ into $h = - \frac{b}{2 a} .$
Find $k,$ the y -coordinate of the vertex, by evaluating $k = f (h) = f (- \frac{b}{2 a}) .$

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).

$\begin{matrix} The horizontal coordinate of the vertex will be at \\ h & = & - \frac{b}{2 a} \\ = & \frac{−6}{2 (2)} \\ = & \frac{6}{4} \\ = & \frac{3}{2} \\ The vertical coordinate of the vertex will be at \\ k & = & f (h) \\ = & f (\frac{3}{2}) \\ = & 2 {(\frac{3}{2})}^{2} - 6 (\frac{3}{2}) + 7 \\ = & \frac{5}{2} \end{matrix}$

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.

\begin{matrix} f (x) & = & a x^{2} + b x + c \\ f (x) & = & 2 x^{2} - 6 x + 7 \end{matrix}

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

f (x) = 2 {(x - \frac{3}{2})}^{2} + \frac{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.

A General Note

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 $f (x) \geq f (- \frac{b}{2 a}),$ or $[f (- \frac{b}{2 a}), \infty);$ the range of a quadratic function written in general form with a negative $a$ value is $f (x) \leq f (- \frac{b}{2 a}),$ or $(- \infty, f (- \frac{b}{2 a})] .$

The range of a quadratic function written in standard form $f (x) = a {(x - h)}^{2} + k$ with a positive $a$ value is $f (x) \geq k;$ the range of a quadratic function written in standard form with a negative $a$ value is $f (x) \leq k .$

How To

Given a quadratic function, find the domain and range.

Identify the domain of any quadratic function as all real numbers.
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.
Determine the maximum or minimum value of the parabola, $k .$
If the parabola has a minimum, the range is given by $f (x) \geq k,$ or $[k, \infty) .$ If the parabola has a maximum, the range is given by $f (x) \leq k,$ or $(- \infty, k] .$

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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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