Answer
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Hint: In order to find an exponential function when the points are given in this question , we need to solve it step by step . Also to solve the given question , we should know the important concepts related to the question like what is an exponential function - An exponential function is a Mathematical function in form \[f{\text{ }}\left( x \right){\text{ }} = {\text{ }}{a^x}\], where “$x$” is a variable and “$a$” is a constant which is called the base of the function and it should be greater than 0 . It is considered to be the most important function in mathematics . The formula of an exponential function in its general form which is going to be applied in the question to solve it , would be - \[{\text{y }} = {\text{ }}a{(b)^x}\]. Just substitute the points given in the formula and solve it , we will be getting our required result .
Complete step-by-step answer:
The question given to us is to find an exponential function given the points are \[\left( { - 1,8} \right)\] and \[\left( {1,2} \right)\].
The formula of an exponential function in its general form which is going to be applied in the question to solve it , would be - \[{\text{y }} = {\text{ }}a{(b)^x}\].
We are given the following points which are true : \[\left( { - 1,8} \right)\] and \[\left( {1,2} \right)\] .
Just substitute the value of x and y in the formula \[{\text{y }} = {\text{ }}a{(b)^x}\]
\[
{\text{8 }} = {\text{ }}a{(b)^{ - 1}} = \dfrac{a}{b} \\
{\text{2 }} = {\text{ }}a{(b)^1} = ab \;
\]
Here we are going to perform some calculations to simplify the given equation by somewhere using equivalent equations and algebraic identities . Equivalent equations are said to be algebraic equations that may have the same solutions if we add or subtract the same number to both sides of an equation - Left hand side or Right hand side of the “equals to” sign. Or we can multiply or divide the same number to both sides of an equation - Left hand side or Right hand side of the equal sign with the method of simplification .
Multiply both the sides of the first equation by ‘b’ to find that –
$8b = a$
Substitute the value of ‘a’ we got from the first equation into the second equation and solve for b:
$
2 = (8b)b \\
\Rightarrow 2 = 8{b^2} \\
\Rightarrow {b^2} = \dfrac{1}{4} \\
\Rightarrow b = \pm \dfrac{1}{2} \;
$
Two equations seem to be possible here. Substitute both the values of b into the either equation to find a. I'll use the second equation for simpler algebra calculations .
As we got value $b = \pm \dfrac{1}{2}$, we will first solve for $b = \dfrac{1}{2}$ and then for $b = - \dfrac{1}{2}$:
If $b = \dfrac{1}{2}$ then =
$
2 = ab \\
\Rightarrow 2 = a \times \dfrac{1}{2} \\
\Rightarrow a = 4 \;
$
Giving us the equation : $y = 4{\left( {\dfrac{1}{2}} \right)^x}$
If $b = - \dfrac{1}{2}$ then =
$
2 = ab \\
\Rightarrow 2 = a \times \left( { - \dfrac{1}{2}} \right) \\
\Rightarrow a = - 4 \;
$
Giving us the equation : $y = - 4{\left( { - \dfrac{1}{2}} \right)^x}$ .
However! In an exponential function, $b > 0$, otherwise many issues arise when trying to graph the function.
The only valid function is $y = 4{\left( {\dfrac{1}{2}} \right)^x}$which is our required answer.
Note: In equivalent equation which have identical solution we can perform multiplication or division by the same non-zero number both L.H.S. and R.H.S. of an equation
In an equivalent equation which has an identical solution we can raise the same odd power to both L.H.S. and R.H.S. of an equation .
Cross check the answer and always keep the final answer simplified .
Remember the properties and apply appropriately .
Complete step-by-step answer:
The question given to us is to find an exponential function given the points are \[\left( { - 1,8} \right)\] and \[\left( {1,2} \right)\].
The formula of an exponential function in its general form which is going to be applied in the question to solve it , would be - \[{\text{y }} = {\text{ }}a{(b)^x}\].
We are given the following points which are true : \[\left( { - 1,8} \right)\] and \[\left( {1,2} \right)\] .
Just substitute the value of x and y in the formula \[{\text{y }} = {\text{ }}a{(b)^x}\]
\[
{\text{8 }} = {\text{ }}a{(b)^{ - 1}} = \dfrac{a}{b} \\
{\text{2 }} = {\text{ }}a{(b)^1} = ab \;
\]
Here we are going to perform some calculations to simplify the given equation by somewhere using equivalent equations and algebraic identities . Equivalent equations are said to be algebraic equations that may have the same solutions if we add or subtract the same number to both sides of an equation - Left hand side or Right hand side of the “equals to” sign. Or we can multiply or divide the same number to both sides of an equation - Left hand side or Right hand side of the equal sign with the method of simplification .
Multiply both the sides of the first equation by ‘b’ to find that –
$8b = a$
Substitute the value of ‘a’ we got from the first equation into the second equation and solve for b:
$
2 = (8b)b \\
\Rightarrow 2 = 8{b^2} \\
\Rightarrow {b^2} = \dfrac{1}{4} \\
\Rightarrow b = \pm \dfrac{1}{2} \;
$
Two equations seem to be possible here. Substitute both the values of b into the either equation to find a. I'll use the second equation for simpler algebra calculations .
As we got value $b = \pm \dfrac{1}{2}$, we will first solve for $b = \dfrac{1}{2}$ and then for $b = - \dfrac{1}{2}$:
If $b = \dfrac{1}{2}$ then =
$
2 = ab \\
\Rightarrow 2 = a \times \dfrac{1}{2} \\
\Rightarrow a = 4 \;
$
Giving us the equation : $y = 4{\left( {\dfrac{1}{2}} \right)^x}$
If $b = - \dfrac{1}{2}$ then =
$
2 = ab \\
\Rightarrow 2 = a \times \left( { - \dfrac{1}{2}} \right) \\
\Rightarrow a = - 4 \;
$
Giving us the equation : $y = - 4{\left( { - \dfrac{1}{2}} \right)^x}$ .
However! In an exponential function, $b > 0$, otherwise many issues arise when trying to graph the function.
The only valid function is $y = 4{\left( {\dfrac{1}{2}} \right)^x}$which is our required answer.
Note: In equivalent equation which have identical solution we can perform multiplication or division by the same non-zero number both L.H.S. and R.H.S. of an equation
In an equivalent equation which has an identical solution we can raise the same odd power to both L.H.S. and R.H.S. of an equation .
Cross check the answer and always keep the final answer simplified .
Remember the properties and apply appropriately .
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