### Instruction: Give an example of a quadratic function that has zeroes at $x=2$ and $x=4$, and that takes the value $6$ when $x=3$.

Enter your answer in the expanded form "ax^2 + bx + c", where a,b,c are replaced by appropriate numbers.### Response: I know that a quadratic function that has zeroes at $x=2$ and $x=4$ must have factors of $(x-2)$ and $(x-4)$ in its factored form.
So I can write the function as $y = a(x-2)(x-4)$, where $a$ is some constant.
To find $a$, I can use the given information that $y=6$ when $x=3$.
Substituting these values into the function, I get $6 = a(3-2)(3-4)$.
Simplifying, I get $6 = a(-1)$, so $a = -6$.
Therefore, the function is $y = -6(x-2)(x-4)$.
To get the expanded form, I need to multiply out the factors.
Using the distributive property, I get $y = -6(x^2 - 6x + 8)$.
Distributing the -6, I get $y = -6x^2 + 36x - 48$.