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If the solutions of plx)= 0 are=14 and 11, which function could be p?

1 Answer

4 votes

We know that a parabolla is given by the form:


p(x)=ax^2+bx+c

We also know that we can factor this equation to get the form:


p(x)=(x-h)(x-k)

Where


p(h)=0\text{ and }p(k)=0

This is, h and k are the roots.

Then we can write:

h = 14

k = 11


p(x)=(x-14)(x-11)

We can also doing the multiplication to give the answer is standard form a parabolla:


\begin{gathered} p(x)=(x-14)(x-11)=x^2-11x-14x+154 \\ p(x)=x^2-25x+154 \end{gathered}

And that's a function as aked in the problem.

User Karan Maru
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