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The equation of a quadratic (or second-degree polynomial) function can be represented in a variety of forms.
The basic quadratic function equation is the following.
||f(x)=x^2||
To modify the curvature, the direction of the opening, or the position of the vertex of a parabola, certain parameters must be added to the equation to create a transformed quadratic function.
There are three equation forms of transformed quadratic functions.
When transforming the basic form, an equation with various parameters is obtained. The standard form indicates how the basic function has been stretched, compressed, reflected, and/or translated.
The standard form of the equation of a quadratic function is
||f(x)=a(x-h)^{2}+k||
where |a,| |h,| and |k| are real numbers that act as parameters.
The parameter |a| is always non-zero.
The parameters |h| and |k| represent the |x|- and |y|-coordinates of the vertex, respectively.
The parameter |a| determines the orientation (i.e., the direction of the opening) of the parabola and the steepness of its curvature.
If the value of |-k/a| is negative, the function does not have any zeros.
Changing these parameters causes a change in the parabola.
A quadratic function’s equation can also be written in its factored form.
|f(x)=a(x-x_{1})(x-x_{2}),|
where |a,| |x_1,| and |x_2| are real numbers that act as parameters.
The parameter |a| is always non-zero.
The parameter |a| determines the orientation (i.e., the direction of the opening) of the parabola and the steepness of its curvature.
The parameters |x_1| and |x_2| represent the zeros of the quadratic function.
If the function does not have any zeros, its equation cannot be expressed using the factored form.
Modifying the parameters causes a change in the parabola.
A quadratic function’s equation can also be written in general form.
|f(x)=ax^{2}+bx+c,|
where |a,| |b,| and |c| are real numbers that act as parameters and |a| is always non-zero.
The general form is the expanded form of the standard and factored form.
The parameter |a| determines the orientation (i.e., direction of the opening) of the parabola and the steepness of its curvature.
The parameter |c| represents the |y|-intercept of the quadratic function.
If the value of the discriminant |b^2-4ac| is negative, the function does not have any zeroes.
Modifying the parameters causes a change in the parabola.
The parameter |a| has the same value whether the equation is expressed in general, standard, or factored form.
To switch from standard form to general form, it suffices to algebraically expand the function’s equation.
Consider the following quadratic function’s equation in standard form: |f(x)=3(x-4)^{2}+5.|
By expanding:
|f(x)=3(x-4)(x-4)+5|
|f(x)=3(x^{2}-4x-4x+16)+5|
|f(x)=3(x^{2}-8x+16)+5|
|f(x)=3x^{2}-24x+48+5|
|f(x)=3x^{2}-24x+53|
The general form of the function’s equation is |f(x)=3x^2-24x+53.|
|(x \pm y)^2 {\color{Red} \neq} (x^2 \pm y^2)|
Therfore, |(x \pm y)^2 {\color{Blue} = } (x \pm y)(x \pm y).|
Next, the product between the two binomials must be calculated.
To switch from factored form to general form, it suffices to algebraically expand the function’s equation.
Consider the following quadratic function’s equation in factored form: |f(x)=4(x-2)(x+7).|
By expanding:
|f(x)=4[x^{2}+7x-2x-14]|
|f(x)=4[x^{2}+5x-14]|
|f(x)=4x^{2}+20x-56|
The general form of the equation of the function is |f(x)=4x^2+20x-56.|
Since the value of the parameter |a| is the same regardless of the equation’s form, only the values of the parameters |h| and |k| must be found.
To find parameter |h| and |k| values, we generally use the formulas
|(h,k) = \displaystyle \left(- \frac{b}{2a}, \frac{4ac-b^2}{4a} \right).|
Note: The two formulas are obtained from the general form |ax^2+bx+c| using the completing the square factorization method.
From the vertex formula |(h,k)|
Consider the following equation in its general form:
|f(x)=3x^{2}-24x+53|
It is necessary to identify the general form’s parameters:
|a=3, b=-24, c=53|
With these values, the values of |h| and |k| can be calculated:
|h=\displaystyle -\frac{b}{2a}=-\frac{(-24)}{2(3)}=\frac{24}{6}=4|
|k=\displaystyle \frac{4ac-b^{2}}{4a}=\frac{4(3)(53)-(-24)^{2}}{4(3)}=\frac{636-576}{12}=\frac{60}{12}=5|
The standard form of the equation of the function is |f(x)=3(x-4)^{2}+5.|
||\begin{align}h&= -\dfrac{b}{2a}\\&=-\dfrac{-24}{2(3)}\\&=\dfrac{24}{6}\\&=4\end{align}||
||\begin{align}k&= \dfrac{4ac-b^{2}}{4a}\\&=\dfrac{4(3)(53)-(-24)^{2}}{4(3)}\\&=\dfrac{636-576}{12}\\&=5\end{align}||
La forme canonique de la fonction est |f(x)=3(x-4)^{2}+5.|
Using the completing the square method
Take the example above and transform the general equation into standard form using the method of completing the square:
|f(x)=3x^{2}-24x+53|
Factor out the greatest common factor so that the coefficient in front of |x^2| is 1
|\displaystyle f(x)=3(x^{2}-8x+\frac{53}{3})|
Add and subtract the term |\displaystyle \left(\frac{b}{2}\right)^{2}|
|\displaystyle f(x)=3(x^{2}-8x{\color{red}+16}+\frac{53}{3}{\color{red}-16})|
Complete the square
|\displaystyle f(x)=3\left((x-4)^{2}+\frac{53}{3}-16\right)|
Simplify
|\displaystyle f(x)=3\left((x-4)^{2}+\frac{5}{3}\right)|
|\displaystyle f(x)=3(x-4)^{2}+3\times\frac{5}{3}|
|\displaystyle f(x) = 3 (x-4)^2 + 5|
The standard form of the function’s equation is |\displaystyle f(x)=3(x-4)^{2}+5.|
The function’s zeroes must be calculated by replacing |f(x)| with |0| or using the following formula.
||x_{1,2}=h \pm \sqrt{-\frac{k}{a}}||
Consider the following quadratic function’s equation in standard form: |f(x)=2(x-1)^2-8.|
First, calculate the zeroes using the following formula.
||\displaystyle x_{1,2}= h \pm \sqrt{-\frac{k}{a}} = 1 \pm \sqrt{-\frac{-8}{2}}||
Next, separate the formula into two parts: one using the + and the other using the -. |x_1| will be a zero and |x_2| will be the other zero.
| \displaystyle x_1 = 1 + 2 = 3|
|\displaystyle x_2 = 1-2 = -1|
The factored form of the function’s equation is |f(x)=2(x-3)(x+1).|
The function’s zeros must be calculated using the quadratic formula:
||\displaystyle x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}.||
Consider the following quadratic function’s equation in general form: |f(x)=2x^2-4x-6.|
The zeroes are calculated using the following formula.
||\displaystyle x_{1,2}= \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2-4 (2) (-6)}}{2 (2)} = \frac{4 \pm \sqrt{64}}{4}||
Next, separate the formula into two parts: one using the + and the other using the -. |x_1| will be a zero and |x_2| will be the other zero.
|\displaystyle x_1 = \frac{4 + 8}{4} =3|
|\displaystyle x_2 = \frac{4-8}{4}=-1|
The factored form of the function’s equation is |f(x)=2(x-3)(x+1).|
The value of the parameter |h| is calculated using the midpoint formula:
||h= \frac{x_1 + x_2}{2}|| where |x_1| and |x_2| are the two zeroes.
Calculate |k,| by replacing |x| in the equation with the value |h| found previously.
Consider the following quadratic function’s equation in factored form: |f(x)=3(x+1)(x-2).|
Calculate |h| with the midpoint formula.
||\displaystyle h = \frac{x_1+x_2}{2} = \frac{-1+2}{2} = \frac{1}{2}||
Replace |x| in the equation with the value |h.| Therefore, the value of |k| is obtained.
|\displaystyle f(x)=3(\frac{1}{2}+1)(\frac{1}{2}-2)|
|\displaystyle f(x) = -\frac{27}{4}|
So, |k= \displaystyle -\frac{27}{4}.|
The standard form of the equation is |\displaystyle f(x)=3\left(x-\frac{1}{2}\right)^2 - \frac{27}{4}.|
The equation can be verified by taking the form obtained at the end and transforming it back to its initial form.
Where do the formulas |\displaystyle h = - \frac{b}{2a}| and |\displaystyle k = \frac{4ac-b^2}{4a}| come from? Also, why is the value of parameter |a| the same whether the equation is in general form or standard form?
The two questions can be answered by completing the square.
However, it is possible to use another clever method to get there!
Expand the standard form:
|a(x-h)^2+k= a(x-h)(x-h) + k|
|a(x-h)(x-h)+k = a(x^2-2xh+h^2)+k|
|a(x^2-2xh+h^2) + k = ax^2 - 2axh + ah^2 + k|
Next, compare the terms of the same degree between the standard form and the general form ( |ax^2+bx+c| ).
|ax^2=ax^2 \Rightarrow a=a|
|\displaystyle -2axh = bx \Rightarrow h = - \frac{b}{2a}|
|\displaystyle ah^2+k = c|
Replace |h| with its value.
|\displaystyle a\left( - \frac{b}{2a} \right)^2 + k =c|
|\displaystyle \frac{ab^2}{4a^2} + k = c|
|\displaystyle \frac{b^2}{4a} + k = c| |\displaystyle k = c - \frac{b^2}{4a} \Rightarrow k = \frac{4ac-b^2}{4a}|
Therefore, the desired result is obtained.