
How to find the inverse function for a quadratic equation\[?\]
Answer
542.7k+ views
Hint:This question describes the operation of addition/ subtraction/ multiplication/ division. To solve this type of question we have to assume one equation in the form of a quadratic equation. Finally, we have to find the value of\[y\]from the quadratic equation. Also, we need to know the multiplication process with the involvement of square and square terms.
Complete step by step solution:
The given question is, we have to find the inverse function for a quadratic equation.
To solve the given question, we have to assume one equation in the form of a quadratic equation as follows, The basic form of a quadratic equation is
\[a{x^2} + bx + c = y\]\[ \to \left( 1 \right)\]
We assume, \[f\left( x \right) = {x^2} - 6x + 2\]
The term\[f(x)\]can be replaced by\[y\]. So, we get
\[y = {x^2} - 6x + 2 \to \left( 2 \right)\]
To solve the above equation we add and subtract\[{3^2}\]with the equation. So, the
equation\[\left( 2 \right)\]becomes,
\[y = {x^2} - 6x + 2 + {3^2} - {3^2}\]
The above equation can also be written as.
\[y = {x^2} - 6x + {3^2} + 2 - {3^2}\]
\[y = \left( {{x^2} - 6x + {3^2}} \right) + 2 - {3^2}\]\[ \to \left( 3 \right)\]
In the above equation, we have\[({x^2} - 6x + {3^2})\]. When it is compared to an algebraic formula we get,
\[\left( {{a^2} - 2ab + {b^2}} \right) = {\left( {a - b} \right)^2}\]
\[\left( {{x^2} - 2 \times x \times 3 + {3^2}} \right) = {\left( {x - 3} \right)^2}\]
So, we get
\[\left( {{x^2} - 6x + {3^2}} \right) = {\left( {x - 3} \right)^2}\]
Let’s substitute the above value in the equation\[\left( 3 \right)\], we get
\[y = {\left( {x - 3} \right)^2} + 2 - 9\]
\[y = {\left( {x - 3} \right)^2} - 7\]
For finding the inverse function of the above quadratic equation we have to
replace\[y\]with\[x\]and\[x\]with\[y\]. So, we get
\[y = {(x - 3)^2} - 7\]
\[ \downarrow \] \[ \downarrow \]
\[x = {(y - 3)^2} - 7\] (Inverse form)
Let’s solve the above equation,
\[x = {(y - 3)^2} - 7\]
It also can be written as
\[x + 7 = {(y - 3)^2}\]
Take square root on both sides of the above equation, we get
\[\left( {y - 3} \right) = \pm \sqrt {x + 7} \to \left( 4 \right)\]
Let’s find the\[y\]value from the above equation
\[y = 3 \pm \sqrt {x + 7} \to \left( 5 \right)\]
So, the final answer is, the inverse function of \[{x^2} - 6x + 2\]is\[y = 3 \pm \sqrt {x + 7} \]. By using the above-mentioned process we can find the inverse function of any quadratic equation.
Note: In this type of question if no equation is given we have to assume an equation in the basic form of a quadratic equation.
To make easy calculation we would try to convert the equation in the form of algebraic formulae like\[{\left( {a - b} \right)^2}\],\[{\left( {a + b} \right)^2}\],\[\left( {{a^2} - {b^2}} \right)\], etc.
To find the inverse function we have to replace the\[x\] term with\[y\]term and \[y\]term with\[x\].
Complete step by step solution:
The given question is, we have to find the inverse function for a quadratic equation.
To solve the given question, we have to assume one equation in the form of a quadratic equation as follows, The basic form of a quadratic equation is
\[a{x^2} + bx + c = y\]\[ \to \left( 1 \right)\]
We assume, \[f\left( x \right) = {x^2} - 6x + 2\]
The term\[f(x)\]can be replaced by\[y\]. So, we get
\[y = {x^2} - 6x + 2 \to \left( 2 \right)\]
To solve the above equation we add and subtract\[{3^2}\]with the equation. So, the
equation\[\left( 2 \right)\]becomes,
\[y = {x^2} - 6x + 2 + {3^2} - {3^2}\]
The above equation can also be written as.
\[y = {x^2} - 6x + {3^2} + 2 - {3^2}\]
\[y = \left( {{x^2} - 6x + {3^2}} \right) + 2 - {3^2}\]\[ \to \left( 3 \right)\]
In the above equation, we have\[({x^2} - 6x + {3^2})\]. When it is compared to an algebraic formula we get,
\[\left( {{a^2} - 2ab + {b^2}} \right) = {\left( {a - b} \right)^2}\]
\[\left( {{x^2} - 2 \times x \times 3 + {3^2}} \right) = {\left( {x - 3} \right)^2}\]
So, we get
\[\left( {{x^2} - 6x + {3^2}} \right) = {\left( {x - 3} \right)^2}\]
Let’s substitute the above value in the equation\[\left( 3 \right)\], we get
\[y = {\left( {x - 3} \right)^2} + 2 - 9\]
\[y = {\left( {x - 3} \right)^2} - 7\]
For finding the inverse function of the above quadratic equation we have to
replace\[y\]with\[x\]and\[x\]with\[y\]. So, we get
\[y = {(x - 3)^2} - 7\]
\[ \downarrow \] \[ \downarrow \]
\[x = {(y - 3)^2} - 7\] (Inverse form)
Let’s solve the above equation,
\[x = {(y - 3)^2} - 7\]
It also can be written as
\[x + 7 = {(y - 3)^2}\]
Take square root on both sides of the above equation, we get
\[\left( {y - 3} \right) = \pm \sqrt {x + 7} \to \left( 4 \right)\]
Let’s find the\[y\]value from the above equation
\[y = 3 \pm \sqrt {x + 7} \to \left( 5 \right)\]
So, the final answer is, the inverse function of \[{x^2} - 6x + 2\]is\[y = 3 \pm \sqrt {x + 7} \]. By using the above-mentioned process we can find the inverse function of any quadratic equation.
Note: In this type of question if no equation is given we have to assume an equation in the basic form of a quadratic equation.
To make easy calculation we would try to convert the equation in the form of algebraic formulae like\[{\left( {a - b} \right)^2}\],\[{\left( {a + b} \right)^2}\],\[\left( {{a^2} - {b^2}} \right)\], etc.
To find the inverse function we have to replace the\[x\] term with\[y\]term and \[y\]term with\[x\].
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

