Question 391614
The general form for an exponential function is:
{{{f(x) = a*b^x}}}
where "a" and "b" are non-zero constants. So the task before us is to figure out the specific "a" and "b" that will cause the graph to pass through the two given points.<br>
If the graph passes through (0, 4) then when we use an input of 0 for the function we should get 4 as the output. IOW: f(0) = 4. Substituting 0 and 4 into the general equation we get:
{{{4 = a*b^0}}}
Since {{{b^0}}} equals 1 this becomes:
4 = a*1
or
4 = a
Already we have our "a".<br>
If the graph passes through (-2, 100) then when we use an input of -2 for the function we should get 100 as the output. IOW: f(-2) = 100. Substituting -2 and 100 (and our "a" (4) into the general equation we get:
{{{100 = (4)*b^(-2)}}}
Dividing both sides by 4 we get:
{{{25 = b^(-2)}}}
Next we need to find a way to change the exponent on "b" to a 1. The way to do this is to raise both sides of the equation to the reciprocal of -2 power. The reasons we do this are:<ul><li>Raising a power to a power means we will multiply the exponents.</li><li>When multiplying reciprocals the answer is <i>always</i> a 1!</li></ul>
The reciprocal of -2 is -1/2. So now we have:
{{{(25)^((-1/2)) = (b^(-2))^((-1/2))}}}
On the right side we get {{{b^1}}} or "b" just as we planned. All we have to do is simplify the left side. If you have trouble with negative and/or fraction exponents I find it helpful to factor the exponent in a certain way:<ul><li>If the exponent is negative, factor out a -1.</li><li>If the exponent is fractional and the numerator is not a 1, factor out the numerator, (For example, factor an exponent like {{{3/4}}} into {{{3*(1/4)}}}).</li></ul>
Factoring our exponent this way we get:
{{{(25)^(((-1)*(1/2))) = b}}}
Looking at the factors of the exponent we see a -1 and a 1/2. The -1 (as an exponent) tells us that a reciprocal will be found. And the 1/2 (as an exponent) means a square root will be found. And since multiplication is Commutative, we can do these operations <i>in any order we choose!</i> Since finding a square root of 25 seems easier than the reciprocal, I choose to start with that. Then we will finish with a reciprocal:
{{{(25)^(((1/2)*(-1))) = b}}}
{{{(25^((1/2)))^(-1) = b}}}
{{{(5)^(-1) = b}}}
{{{1/5 = b}}}<br>
Now that we have our "a" and "b" we can write the desired function:
{{{f(x) = 4*(1/5)^x}}}
or (if you prefer decimals):
{{{f(x) = 4*(0.2)^x}}}