Your beginning function does not have to look exactly like y=ax2+c{\displaystyle y=ax^{2}+c}. As long as you can look at it and see that the function consists only of x2{\displaystyle x^{2}} terms and constant numbers, you will be able to use this method. For example, suppose you begin with the equation, 2y−6+x2=y+3x2−4{\displaystyle 2y-6+x^{2}=y+3x^{2}-4}. A quick examination of this equation shows that there are no terms of x{\displaystyle x} to the first power. This equation is a candidate for this method to find an inverse function.
Taking the sample equation, 2y−6+x2=y+3x2−4{\displaystyle 2y-6+x^{2}=y+3x^{2}-4}, the y-terms can be consolidated on the left by subtracting a y from both sides. The other terms can be consolidated on the right by adding 6 to both sides and subtracting x^2 from both sides. The resulting equation will be y=2x2+2{\displaystyle y=2x^{2}+2}.
Consider the sample equation y=2x2+2{\displaystyle y=2x^{2}+2}. There is no limitation on allowable values of x for this equation. However, you should recognize that this is the equation of a parabola, centered at x=0, and a parabola is not a function because it does not consist of a one-to-one mapping of x and y values. To limit this equation and make it a function, for which we can find an inverse, we must define the domain as x≥0. The range is similarly limited. Notice that the first term, 2x2{\displaystyle 2x^{2}}, will always be positive or 0, for any value of x. When the equation then adds +2, the range will be any values y≥2. Defining the domain and range at this early stage is necessary. You will use these definitions later in defining the domain and range of the inverse function. In fact, the domain of the original function will become the range of the inverse function, and the range of the original will become the domain of the inverse. [2] X Research source
Working with the sample equation y=2x2+2{\displaystyle y=2x^{2}+2}, this inversion step will result in the new equation of x=2y2+2{\displaystyle x=2y^{2}+2}. An alternate format is to replace the y terms with x, but replace the x terms with either y−1{\displaystyle y^{-}1} or f(x)−1{\displaystyle f(x)^{-}1} to indicate the inverse function.
x=2y2+2{\displaystyle x=2y^{2}+2} (original starting point) x−2=2y2{\displaystyle x-2=2y^{2}} (subtract 2 from both sides) x−22=y2{\displaystyle {\frac {x-2}{2}}=y^{2}} (divide both sides by 2) ±x−22=y{\displaystyle {\sqrt {\frac {x-2}{2}}}=y} (square root of both sides; remember that the square root results in both positive and negative possible answers)
Examine the sample equation solution of ±x−22=y{\displaystyle {\sqrt {\frac {x-2}{2}}}=y}. Because the square root function is not defined for any negative values, the term x−22{\displaystyle {\frac {x-2}{2}}} must always be positive. Therefore, allowable values of x (the domain) must be x≥2. Using that as the domain, the resulting values of y (the range) are either all values y≥0, if you take the positive solution of the square root, or y≤0, if you select the negative solution of the square root. Recall that you originally defined the domain as x≥0, in order to be able to find the inverse function. Therefore, the correct solution for the inverse function is the positive option. Compare the domain and range of the inverse to the domain and range of the original. Recall that for the original function, y=2x2+2{\displaystyle y=2x^{2}+2}, the domain was defined as all values of x≥0, and the range was defined as all values y≥2. For the inverse function, now, these values switch, and the domain is all values x≥2, and the range is all values of y≥0.
As a sample, select the value x=1 to place in the original equation y=2x2+2{\displaystyle y=2x^{2}+2}. This gives the result y=4. Next, place that value of 4 into the inverse function x−22=y{\displaystyle {\sqrt {\frac {x-2}{2}}}=y}. This does give the result of y=1. You can conclude that your inverse function is correct.
The first thing to notice is the value of the coefficient a. If a>0, then the equation defines a parabola whose ends point upward. If a<0, the equation defines a parabola whose ends point downward. Notice that a≠0. If it did, then this would be a linear function and not quadratic.
Notice that this standard format consists of a perfect square term, (x−h)2{\displaystyle (x-h)^{2}}, which is then adjusted by the other two elements a and k. To get to this perfect square form, you will need to create certain conditions in your quadratic equation.
To complete the square, you will be working in reverse. You will start with x2{\displaystyle x^{2}} and some second x-term. From the coefficient of that term, which you can define as “2b,” you will need to find b2{\displaystyle b^{2}}. This will require a combination of dividing by two and then squaring that result.
For example, consider the quadratic function f(x)=2x2+6x−4{\displaystyle f(x)=2x^{2}+6x-4}. You must simplify this by dividing all terms by 2, to yield the resulting function f(x)=2(x2+3x−2){\displaystyle f(x)=2(x^{2}+3x-2)}. The coefficient 2 will remain outside of the parentheses and will be part of your final solution. If all terms are not multiples of a, you will wind up with fractional coefficients. For example, the function f(x)=3x2−2x+6{\displaystyle f(x)=3x^{2}-2x+6} will simplify to f(x)=3(x2−2x3+2){\displaystyle f(x)=3(x^{2}-{\frac {2x}{3}}+2)}. Work carefully with the fractions as necessary.
For example, if the first two terms of your quadratic function are x2+3x{\displaystyle x^{2}+3x}, you will find the needed third term by dividing 3 by 2, which gives the result 3/2, and then squaring that, to get 9/4. The quadratic x2+3x+9/4{\displaystyle x^{2}+3x+9/4} is a perfect square. As another example, suppose your first two terms are x2−4x{\displaystyle x^{2}-4x}. Half of the middle term is -2, and then you square that to get 4. The resulting perfect square quadratic is x2−4x+4{\displaystyle x^{2}-4x+4}.
Suppose you have the function f(x)=x2−4x+9{\displaystyle f(x)=x^{2}-4x+9}. As noted above, you will use the first two terms to work on completing the square. Using the middle term of -4x, you will generate a third term of +4. Add and subtract 4 to the equation, in the form f(x)=(x2−4x+4)+9−4{\displaystyle f(x)=(x^{2}-4x+4)+9-4}. The parentheses are placed just to define the perfect square quadratic that you are creating. Notice the +4 inside the parentheses and the -4 on the outside. Simplify the numbers to give the result of f(x)=(x2−4x+4)+5{\displaystyle f(x)=(x^{2}-4x+4)+5}.
Notice that for this function, a=1, h=2, and k=5. The value of writing the equation in this form is that a, being positive, tells you that the parabola points upward. The values of (h,k) tell you the apex point at the bottom of the parabola, if you wanted to graph it.
Continue working with the sample function f(x)=(x−2)2+5{\displaystyle f(x)=(x-2)^{2}+5}. Because this is in standard format, you can identify the apex point as x=2, y=5. Thus, to avoid the symmetry, you will only work with the right-side of the graph, and set the domain as all values x≥2. Inserting the value x=2 into the function gives the result of y=5. You can see that the values of y will increase as x increases. Therefore the range of this equation is y≥5.
Continue to work with the function f(x)=(x−2)2+5{\displaystyle f(x)=(x-2)^{2}+5}. Insert x in place of f(x), and insert y (or f(x), if you prefer) in place of x. This will yield the new function x=(y−2)2+5{\displaystyle x=(y-2)^{2}+5}.
x=(y−2)2+5{\displaystyle x=(y-2)^{2}+5} (original starting point) x−5=(y−2)2{\displaystyle x-5=(y-2)^{2}} (subract 5 from both sides) ±x−5=y−2{\displaystyle {\sqrt {x-5}}=y-2} (square root of both sides; remember that the square root results in both positive and negative possible answers) ±x−5+2=y{\displaystyle {\sqrt {x-5}}+2=y} (add 2 to both sides)
Examine the sample equation solution of ±x−5+2=y{\displaystyle {\sqrt {x-5}}+2=y}. Because the square root function is not defined for any negative values, the term x−5{\displaystyle {x-5}} must always be positive. Therefore, allowable values of x (the domain) must be x≥5. Using that as the domain, the resulting values of y (the range) are either all values y≥2, if you take the positive solution of the square root, or y≤2 if you select the negative solution of the square root. Recall that you originally defined the domain as x≥2, in order to be able to find the inverse function. Therefore, the correct solution for the inverse function is the positive option. Compare the domain and range of the inverse to the domain and range of the original. Recall that for the original function the domain was defined as all values of x≥2, and the range was defined as all values y≥5. For the inverse function, now, these values switch, and the domain is all values x≥5, and the range is all values of y≥2.
As a sample, select the value x=3 to place in the original equation f(x)=x2−4x+9{\displaystyle f(x)=x^{2}-4x+9}. This gives the result y=6. Next, place that value of 6 into the inverse function x−5+2=y{\displaystyle {\sqrt {x-5}}+2=y}. This does give the result of y=3, which is the number you started with. You can conclude that your inverse function is correct.
The Quadratic Formula is x=[-b±√(b^2-4ac)]/2a. Notice that the Quadratic Formula will result in two possible solutions, one positive and one negative. You will make this selection based on defining the domain and range of the function.
For this section of this article, use the sample equation f(x)=x2+2x−3{\displaystyle f(x)=x^{2}+2x-3}.
Using the working equation f(x)=x2+2x−3{\displaystyle f(x)=x^{2}+2x-3}, this will give the result x=y2+2y−3{\displaystyle x=y^{2}+2y-3}.
For the sample equation, to get the left side equal to 0, you must subtract x from both sides of the equation. This will give the result 0=y2+2y−3−x{\displaystyle 0=y^{2}+2y-3-x}.
Let y2=ax2{\displaystyle y^{2}=ax^{2}}. Therefore, x=1 Let 2y=bx{\displaystyle 2y=bx}. Therefore, b=2 Let (−3−x)=c{\displaystyle (-3-x)=c}. Therefore, c=(-3-x)
x=[-b±√(b^2-4ac)]/2a x=(-2)±√((-2)^2-4(1)(-3-x)) / 2(1) x=((-2)±√(4+12+4x))/2 x=(-2±√(16+4x))/2 x=(-2±√(4)(4+x))/2 x=-2±2√(4+x))/2 x=-1±√(4+x) f-inverse = -1±√(4+x) (This final step is possible because you earlier put x in place of the f(x) variable. )
f−1=−1+4+x{\displaystyle f^{-1}=-1+{\sqrt {4+x}}} f−1=−1−4+x{\displaystyle f^{-1}=-1-{\sqrt {4+x}}}
Using the original function f(x)=x2+2x−3{\displaystyle f(x)=x^{2}+2x-3}, choose x=-2. This will give the result of y=-3. Now put the value of x=-3 into the inverse function, f−1=−1−4+x{\displaystyle f^{-1}=-1-{\sqrt {4+x}}}. This turns out the result of -2, which is indeed the value that you started with. Therefore, your definition of the inverse function is correct.
