Quadratic functions are functions that can be written in the form
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$\,y = ax^2 + bx + c\,$ for $\,a\ne 0\,$.
Every quadratic function graphs as a parabola with directrix parallel to the $\,x$-axis.
The graphs of quadratic functions can have three different $\,x$-intercept situations, as shown below:
no
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$\,x$-intercepts;
exactly one $\,x$-intercept;
two different $\,x$-intercepts.
no [beautiful math coming... please be patient] $\,x\,$-intercepts | exactly one $\,x\,$-intercept | two different $\,x\,$-intercepts |
Points on the
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$\,x$-axis have their $\,y$-value equal to zero.
Thus, to find the $\,x$-intercepts for any curve, you set $\,y\,$
equal to zero and solve for $\,x\,$.
In particular, to find the $\,x$-intercepts of a quadratic function
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$\,y = ax^2 + bx + c\,$ ($\,a\ne 0\,$),
it is necessary to solve the equation
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$\,ax^2 + bx + c = 0\,$.
The formula that gives the solutions to this equation is called the quadratic formula, and is derived next:
Using the technique of
completing the square,
the equation is transformed to the form
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$\,z^2 = k\,$, as follows:
[beautiful math coming... please be patient] $\displaystyle ax^2 + bx + c = 0$ | original equation |
[beautiful math coming... please be patient] $\displaystyle ax^2 + bx = -c$ | subtract $\,c\,$ from both sides |
[beautiful math coming... please be patient] $\displaystyle x^2 + \frac{b}{a}x = -\frac{c}{a}$ | divide both sides by $\,a\ne 0\,$; the technique of completing the square only works when the coefficient of the squared term is $\,1\,$ |
[beautiful math coming... please be patient] $\displaystyle x^2 + \frac{b}{a}x + {(\frac{b}{2a})}^2 = -\frac{c}{a} + \frac{b^2}{4a^2}$ | notice that $\,\frac{b/a}{2} = \frac{b}{a}\div 2 = \frac{b}{a}\cdot \frac{1}{2} = \frac{b}{2a}\,$; add [beautiful math coming... please be patient] $\,{(\frac{b}{2a})}^2 = \frac{b^2}{4a^2}\,$ to both sides; this is the correct number to add to turn the left side into a perfect square |
[beautiful math coming... please be patient] $\displaystyle{\left(x + \frac{b}{2a}\right)}^2 = -\frac{c}{a}\cdot \frac{4a}{4a} + \frac{b^2}{4a^2}$ | rewrite left side as a perfect square; get a common denominator on the right |
[beautiful math coming... please be patient] $\displaystyle{\left(x + \frac{b}{2a}\right)}^2 = \frac{b^2 - 4ac}{4a^2}$ | add fractions, simplify |
[beautiful math coming... please be patient] $\displaystyle x + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}}$ | the equation $\,z^2 = k\,$ has solutions
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$\,z=\pm\sqrt{k}\ $; recall that $\,z=\pm\sqrt{k}\,$ is just a convenient shorthand for $\,(z = \sqrt{k}\ \ \text{or}\ \ z = -\sqrt{k})$ |
[beautiful math coming... please be patient] $\displaystyle x + \frac{b}{2a} = \pm \frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}$ | use the rule [beautiful math coming... please be patient] $\displaystyle \,\sqrt{\frac{A}{B}} = \frac{\sqrt{A}}{\sqrt{B}}\,$ |
[beautiful math coming... please be patient] $\displaystyle x + \frac{b}{2a} = \pm \frac{\sqrt{b^2-4ac}}{\sqrt{4}\sqrt{a^2}}$ | use the rule [beautiful math coming... please be patient] $\displaystyle \,\sqrt{AB} = \sqrt{A}\cdot\sqrt{B}\,$ |
[beautiful math coming... please be patient] $\displaystyle x + \frac{b}{2a} = \pm \frac{\sqrt{b^2-4ac}}{2|a|}$ | use the rule $\,\sqrt{A^2} = |A|\,$ |
Notice that if [beautiful math coming... please be patient] $\,a\gt 0\,$, then $\,|a|=a\,$, so the right side of the previous equation becomes [beautiful math coming... please be patient] $\,\pm\frac{\sqrt{b^2-4ac}}{2a}\,$.
Also, if $\,a\lt 0\,$, then $\,|a|=-a\,$, so the right side becomes [beautiful math coming... please be patient] $\,\pm\frac{\sqrt{b^2-4ac}}{2(-a)} = \mp\frac{\sqrt{b^2-4ac}}{2a}\,$.
In both cases, the same two values for the right side result.
Thus, we continue with a simplified right side:
[beautiful math coming... please be patient] $\displaystyle x + \frac{b}{2a} = \pm \frac{\sqrt{b^2-4ac}}{2a}$ | the plus/minus allows elimination of the absolute value |
[beautiful math coming... please be patient] $\displaystyle x=-\frac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$ | add $\displaystyle\,-\frac{b}{2a}\,$ to both sides |
[beautiful math coming... please be patient] $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ | add fractions |
In summary, we have:
The expression
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$\,b^2 - 4ac\,$ that appears
under the square root in the quadratic formula
is critical in determining the nature of the solutions
to the quadratic equation
$\,ax^2 + bx + c = 0\,$, as follows:
Thus, the expression
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$\,b^2 - 4ac\,$ helps us to
discriminate between the various types of solutions to a quadratic equation,
and the various $\,x\,$-intercept situations for
a quadratic function.
Thus, we have the following definition: