Question

select the correct answer.
which function is nonlinear?
a. $9y + 3 = 0$
b. $y - 4x = 1$
c. $y = 2 + 6x^4$
d. $x - 2y = 7$
e. $\frac{x}{y} + 1 = 2$

Explanation:

Step1: Recall Linear Function Definition

A linear function has the form \( y = mx + b \) (or can be rewritten in this form), where the highest power of \( x \) (or other variables) is 1, and variables are not in denominators, exponents other than 1, etc.

Step2: Analyze Option A

For \( 9y + 3 = 0 \), solve for \( y \): \( 9y=-3 \) → \( y = -\frac{3}{9}=-\frac{1}{3} \). This is a horizontal line (constant function), which is linear (power of \( x \) is 0, still linear as it's a special case of \( y = mx + b \) with \( m = 0 \)).

Step3: Analyze Option B

For \( y - 4x = 1 \), solve for \( y \): \( y = 4x + 1 \). This is in \( y = mx + b \) form with \( m = 4 \), \( b = 1 \). Linear.

Step4: Analyze Option C

For \( y = 2 + 6x^4 \), the term \( x^4 \) has a power of 4 (greater than 1). So this function is nonlinear (polynomial of degree 4, not linear).

Step5: Analyze Option D

For \( x - 2y = 7 \), solve for \( y \): \( -2y=-x + 7 \) → \( y=\frac{1}{2}x-\frac{7}{2} \). This is in \( y = mx + b \) form. Linear.

Step6: Analyze Option E

For \( \frac{x}{y}+1 = 2 \), rewrite: \( \frac{x}{y}=1 \) → \( x = y \) (or \( y = x \) when \( y
eq0 \)). Wait, no: \( \frac{x}{y}=1 \) → \( x = y \) (linear). Wait, maybe miscalculation. Wait, \( \frac{x}{y}+1 = 2 \) → \( \frac{x}{y}=1 \) → \( x = y \) (so \( y = x \), linear). But wait, original equation: \( \frac{x}{y}=1 \) implies \( y = x \) (for \( y
eq0 \)), which is linear. But wait, let's check again. Wait, maybe I made a mistake. Wait, the equation \( \frac{x}{y}+1 = 2 \) can be rewritten as \( \frac{x}{y}=1 \) → \( x = y \) (so \( y = x \), linear). But option C has \( x^4 \), which is degree 4, so nonlinear. Wait, but let's confirm option E. Wait, \( \frac{x}{y}=1 \) is equivalent to \( y = x \) (linear), but if we consider the original equation, \( y \) is in the denominator, but when we solve it, we get a linear relation (though \( y
eq0 \), but as a function, if we consider \( y \) as a function of \( x \), \( y = x \) (linear). However, option C has \( x^4 \), which is a higher power, so it's nonlinear.

Answer:

C. \( y = 2 + 6x^4 \)