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Section 11.2 Equations of Quadaratic Functions

Example 11.11. Guiding Example.

Suppose \(f(x)=2(x+1)(x-3)\) and we want to find its roots. From Example 11.4, we know that we can set each part equal to 0 and solve. Of course, 2 is not equal to 0, so that doesn't give us any information about the roots. For the other two parts, we will move the numbers to the other side of the equal sign, to get \(x=-1,3\text{.}\)

Notice what happened in the example above:

  • The \(2\) multiplied out front didn't have anything to do with the roots.

  • The \((x+1)\) and \((x-3)\) told us exactly what the root are, but the signs are flipped because the roots come from setting those equal to 0

Those two key observations give us our strategy for working backwards: finding the equation when we are given the roots and some other information.

Definition 11.12.

If \(f(x)\) is a quadratic function with a root \(x=r\text{,}\) then it has a factor of \((x-r)\text{.}\)

Example 11.13.

Suppose \(f(x)\) has roots \(x=2,-5\text{.}\) Then the factors of \(f(x)\) are \((x-2)\) and \((x-(-5))=(x+5)\text{.}\)

Once we know the factors, we have most of the information to build the full equation. However, we still need to figure out the number multiplied out front, like the \(2\) in the example at the start of this section. To do that, we will use the strategy we saw before with lines, like in Example 7.12.

Example 11.14.

Suppose \(f(x)\) is a quadratic function with roots \(x=-1,3\) and goes through the point \((5,-4)\text{.}\) Let's write the equation of that function.

We already know from the discussion above that \(f(x)\) has factors \((x+1)\) and \((x-3)\text{,}\) so our function looks like this so far:

\begin{equation*} f(x)=A(x+1)(x-3) \end{equation*}

We just need to figure out what \(A\) is. The other piece of information we are given is that it goes through the point \((5,-4)\text{,}\) so just like we did with linear functions, we will plug in \(x=5\) and \(y=-4\) and solve for the number we don't know.

\begin{align*} f(x)\amp=A(x+1)(x-3)\\ -4\amp=A(5+1)(5-3)\\ -4\amp=A(6)(2)\\ -4\amp=A(12)\\ \dfrac{-4}{12}\amp=A\\ -\dfrac{1}{3}\amp=A \end{align*}

Now we can plug that in to get our final answer:

\begin{equation*} f(x)=-\dfrac{1}{3}(x+1)(x-3) \end{equation*}

Checkpoint 11.15.

Suppose \(f(x)\) is a quadratic function with roots \(x=-3,-2\) and \(y\)-intercept at \((0,12)\text{.}\) Write the equation for \(f(x)\text{.}\)
Answer.
\(f(x)=2(x+3)(x+2)\)
Solution.

Right away, we can see that \(f(x)\) has factors \((x+3)\) and \((x+2)\text{,}\) so our function looks like this so far:

\begin{equation*} f(x)=A(x+3)(x+2) \end{equation*}

We just need to figure out what \(A\) is. The other piece of information we are given is that it goes through the point \((0,12)\text{,}\) so we will plug in \(x=0\) and \(y=12\) and solve for the number we don't know.

\begin{align*} f(x)\amp=A(x+3)(x+2)\\ 12\amp=A(0+3)(0+2)\\ 12\amp=A(3)(2)\\ 12\amp=A(6)\\ \dfrac{12}{6}\amp=A\\ 2\amp=A \end{align*}

Now we can plug that in to get our final answer:

\begin{equation*} f(x)=2(x+3)(x+2) \end{equation*}

Example 11.16.

Suppose \(f(x)\) is a quadratic function with one root at \(x=3\) and goes through the point \((1,20)\text{.}\) Let's write the equation of \(f(x)\text{.}\)

Since we know the root, we know the factor is \((x-3)\text{.}\) However, to be a quadratic function, we need to actually have two factors. Since this one has one root, both factors must be the same thing. So, our function looks like this:

\begin{equation*} f(x)=A(x-3)(x-3)=A(x-3)^2 \end{equation*}

From here, we follow the same strategy as before. We know the function goes through the point \((1,20)\text{,}\) so we can plug in \(x=1\) and \(y=20\) and solve for \(A\text{.}\)

\begin{align*} f(x)\amp=A(x-3)^2\\ 20\amp=A(1-3)^2\\ 20\amp=A(-2)^2\\ 20\amp=A(4)\\ 5\amp=A \end{align*}

Now we can plug that in to get our final answer:

\begin{equation*} f(x)=5(x-3)^2 \end{equation*}

Checkpoint 11.17.

Suppose \(f(x)\) is a quadratic function with one root at \(x=-2\) and goes through the point \((2,-2)\text{.}\) Write the equation for \(f(x)\text{.}\)
Answer.
\(f(x)=-\dfrac{1}{8}(x+2)^2\)
Solution.

Since we know the root, we know the factor is \((x+2)\text{.}\) However, to be a quadratic function, we need to actually have two factors. Since this one has one root, both factors must be the same thing. So, our function looks like this:

\begin{equation*} f(x)=A(x+2)(x+2)=A(x+2)^2 \end{equation*}

From here, we follow the same strategy as before. We know the function goes through the point \((2,-2)\text{,}\) so we can plug in \(x=2\) and \(y=-2\) and solve for \(A\text{.}\)

\begin{align*} f(x)\amp=A(x+2)^2\\ -2\amp=A(2+2)^2\\ -2\amp=A(4)^2\\ -2\amp=A(16)\\ \dfrac{-2}{16}\amp=A\\ -\dfrac{1}{8}\amp=A \end{align*}

Now we can plug that in to get our final answer:

\begin{equation*} f(x)=-\dfrac{1}{8}(x+2)^2 \end{equation*}