Question 68·SAT Desmos Regression Practice·Advanced Math

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For $x\ne2$ , let $u=x-2$ . The expression below can be written equivalently as $ax^2+bx+c$ , where $a$ , $b$ , and $c$ are constants.

\frac{(2u^2+3u)^3-(2u^2-3u)^3}{2u^3}

What is the value of $b$ ?

Your answer

The cue is a quadratic form with unknown coefficients: you need the coefficient of $x$ , not a value of $x$ . Because the problem guarantees that form, you can use three distinct allowed inputs in a Desmos regression to find its coefficients. Keep the restriction in view: an input that makes the original denominator zero cannot be used.

Hints

Hint 1
A quadratic has three coefficients in $ax^2+bx+c$ . Three distinct inputs give enough values to determine them. Which nearby inputs can you choose without using the excluded value?
Hint 2
The expression uses $u=x-2$ , so define $u$ as a function of $x$ before entering the fraction. At each chosen $x$ , Desmos can then calculate the matching value of $u$ .
Hint 3
A Desmos regression uses $\sim$ to find unknown coefficients. Match the fraction's outputs to $ax_1^2+bx_1+c$ , then read $b$ under PARAMETERS.

Step-by-step

Approach 1: Fit the quadratic in Desmos
Step 1
Choose three allowed inputs
Type $x_1=[-1,0,1]$ in Desmos. The problem says the expression equals a quadratic $ax^2+bx+c$ , so three distinct allowed inputs determine its three coefficients. All three choices avoid $x=2$ , where the original denominator would be zero.
Step 2
Evaluate the given expression
Add the four function definitions shown, then type $f(x_1)$ . Here $p$ and $q$ name the two terms being cubed, so $f$ is the original fraction with $u=x-2$ . Desmos gives $[351,171,63]$ for the inputs $[-1,0,1]$ , in that order.
Step 3
Read the coefficient of x
Add $f(x_1)\sim ax_1^2+bx_1+c$ . The regression finds the quadratic matching those three outputs; Desmos shows $a=36$ , $b=-144$ , and $c=171$ under PARAMETERS. Since $b$ multiplies $x$ in the requested form, its value is $-144$ . Grid in -144.

Approach 2: See why the cubes leave a quadratic
Step 1
Expand the two cubes
Let $A=2u^2$ and $B=3u$ . The numerator is $(A+B)^3-(A-B)^3$ . Expand each cube, keeping its middle terms:
$(A+B)^3=A^3+3A^2B+3AB^2+B^3$ $(A-B)^3=A^3-3A^2B+3AB^2-B^3.$
Step 2
Cancel matching terms
Subtract the second expansion from the first. The $A^3$ and $3AB^2$ terms cancel, while the other pairs double:
$(A+B)^3-(A-B)^3=6A^2B+2B^3.$
Step 3
Put u back into the fraction
Replace $A$ with $2u^2$ and $B$ with $3u$ in the numerator:
$\frac{(2u^2+3u)^3-(2u^2-3u)^3}{2u^3} =\frac{6(2u^2)^2(3u)+2(3u)^3}{2u^3}.$
The whole numerator is still divided by $2u^3$ .
Step 4
Divide each numerator term
Because $x\ne2$ , $u\ne0$ , so each numerator term can be divided by $2u^3$ . Type the two numerical coefficients shown into Desmos; it gives $36$ and $27$ . Apply the matching exponent subtraction to the powers of $u$ :
$\frac{6(2u^2)^2(3u)+2(3u)^3}{2u^3}=36u^2+27.$
Step 5
Expand the square in x
Substitute the given $u=x-2$ , then square the binomial. Its middle term is negative:
$36u^2+27=36(x-2)^2+27$ $=36(x^2-4x+4)+27.$
Step 6
Identify b
Only $36(-4x)$ produces an $x$ term. Multiply its coefficients:
$36(-4x)=-144x.$
So $b=-144$ , the coefficient of $x$ in $ax^2+bx+c$ . Grid in -144.

Question 68·SAT Desmos Regression Practice·Advanced Math

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