Question 78·Medium·Math

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For the quadratic function $f$ , the table shows three values of $x$ and their corresponding values of $f(x)$ . Which equation defines $f$ ?

$x$	$f(x)$
$-1$	$10$
$0$	$14$
$1$	$20$

When a function is given by a table and the answer choices are specific formulas, use substitution and elimination: first plug in easy values like $x=0$ to quickly find or confirm the constant term, then either solve for remaining coefficients using the other points or, even faster, plug each given $x$ -value into the answer choices and eliminate any that do not produce all the listed $f(x)$ -values. This targeted checking is usually much quicker and less error-prone than trying to guess or memorize forms.

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Start with the easiest x-value

Look at the row where $x=0$ . What does $f(0)=14$ tell you about the constant term $c$ in a quadratic $f(x)=ax^2+bx+c$ ?

Use the remaining points to create equations

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 $b$ .

Solve the system or test choices

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 $f(1)=20$ .

Desmos Guide

Create the table from the problem

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.

Graph each candidate function

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.

Compare function values to the table

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.

Step-by-step Explanation

Use the value at x = 0 to find the constant term

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.

Write equations using the other two points

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

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

\begin{aligned} a(1)^2 + b(1) + 14 &= 20 \\ a + b + 14 &= 20 \\ a + b &= 6 \end{aligned}

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

\begin{aligned} a(-1)^2 + b(-1) + 14 &= 10 \\ a - b + 14 &= 10 \\ a - b &= -4 \end{aligned}

So we have a system of two equations:

$a + b = 6$
$a - b = -4$

Solve for a and b

Add the two equations to eliminate $b$ :

\begin{aligned} (a + b) + (a - b) &= 6 + (-4) \\ 2a &= 2 \\ a &= 1 \end{aligned}

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

\begin{aligned} 1 + b &= 6 \\ b &= 5 \end{aligned}

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

Write the function and match it to a choice

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

f(x)=x^2+5x+14.

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

Question 78·Medium·Math

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Step-by-step Explanation

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Step-by-step Explanation