Answer :
To find a quadratic function with zeros (or roots) at [tex]\( x = 6 \)[/tex] and [tex]\( x = 4 \)[/tex], we start by writing the function in its factored form. If a quadratic function has zeros [tex]\( r_1 \)[/tex] and [tex]\( r_2 \)[/tex], then it can be written as
[tex]$$
f(x) = (x - r_1)(x - r_2).
$$[/tex]
Given [tex]\( r_1 = 6 \)[/tex] and [tex]\( r_2 = 4 \)[/tex], substitute these values to get
[tex]$$
f(x) = (x - 6)(x - 4).
$$[/tex]
Next, we expand the product:
[tex]\[
\begin{aligned}
f(x) &= (x - 6)(x - 4) \\
&= x(x - 4) - 6(x - 4) \\
&= x^2 - 4x - 6x + 24 \\
&= x^2 - 10x + 24.
\end{aligned}
\][/tex]
Thus, the quadratic function is
[tex]$$
f(x) = x^2 - 10x + 24.
$$[/tex]
To verify, substitute the zeros into the function:
1. For [tex]\( x = 6 \)[/tex]:
[tex]$$
f(6) = 6^2 - 10 \cdot 6 + 24 = 36 - 60 + 24 = 0.
$$[/tex]
2. For [tex]\( x = 4 \)[/tex]:
[tex]$$
f(4) = 4^2 - 10 \cdot 4 + 24 = 16 - 40 + 24 = 0.
$$[/tex]
Since both evaluations give 0, the function has the correct zeros.
Therefore, the correct choice is
[tex]$$
f(x) = x^2 - 10x + 24.
$$[/tex]
[tex]$$
f(x) = (x - r_1)(x - r_2).
$$[/tex]
Given [tex]\( r_1 = 6 \)[/tex] and [tex]\( r_2 = 4 \)[/tex], substitute these values to get
[tex]$$
f(x) = (x - 6)(x - 4).
$$[/tex]
Next, we expand the product:
[tex]\[
\begin{aligned}
f(x) &= (x - 6)(x - 4) \\
&= x(x - 4) - 6(x - 4) \\
&= x^2 - 4x - 6x + 24 \\
&= x^2 - 10x + 24.
\end{aligned}
\][/tex]
Thus, the quadratic function is
[tex]$$
f(x) = x^2 - 10x + 24.
$$[/tex]
To verify, substitute the zeros into the function:
1. For [tex]\( x = 6 \)[/tex]:
[tex]$$
f(6) = 6^2 - 10 \cdot 6 + 24 = 36 - 60 + 24 = 0.
$$[/tex]
2. For [tex]\( x = 4 \)[/tex]:
[tex]$$
f(4) = 4^2 - 10 \cdot 4 + 24 = 16 - 40 + 24 = 0.
$$[/tex]
Since both evaluations give 0, the function has the correct zeros.
Therefore, the correct choice is
[tex]$$
f(x) = x^2 - 10x + 24.
$$[/tex]
