Once you know $c$, rewrite the quadratic as $f(x)=ax^2+bx+14$. Then plug in $x=1$ and $x=-1$ using the table values to form equations involving $a$ and $b$.

You can either solve the two equations for $a$ and $b$, or plug $x=-1$ and $x=1$ into each answer choice and see which one produces $f(-1)=10$ and $f(1)=20$.

Enter the three points from the table as a table in Desmos: in one expression line create a table and enter $x$-values $-1$, $0$, and $1$ in the left column and the corresponding $f(x)$-values $10$, $14$, and $20$ in the right column.

On separate lines, type each option as a function, for example y = 3x^2 + 3x + 14, y = 5x^2 + x + 14, y = 9x^2 - x + 14, and y = x^2 + 5x + 14 so that all four graphs appear on the same coordinate plane.

For each function, click on the graph at $x=-1$, $x=0$, and $x=1$ (or use a table of values for each function) and compare the $y$-values to $10$, $14$, and $20$. The correct equation is the one whose graph passes through all three data points from the original table.

A quadratic in standard form looks like $f(x)=ax^2+bx+c$.

From the table, $f(0)=14$. Plugging $x=0$ into $ax^2+bx+c$ gives $f(0)=c$.

So $c=14$, which matches all four answer choices, so we need more information to distinguish them.

Now use the other two points with the general form $f(x)=ax^2+bx+14$.

From $x=1$, $f(1)=20$:

From $x=-1$, $f(-1)=10$:

So we have a system of two equations:

• $a + b = 6$
• $a - b = -4$

Add the two equations to eliminate $b$:

Now plug $a=1$ into $a + b = 6$:

So the quadratic must have $a=1$, $b=5$, and $c=14$.

Substitute $a=1$, $b=5$, and $c=14$ into $f(x)=ax^2+bx+c$ to get

This matches choice D, so the correct answer is $f(x)=x^2+5x+14$.