Question

Which of the following equations shows a linear relationship between $x$ and $y$?

• A. $y=x+x^{\frac{1}{4}}$
• B. $y=8x+3$
• C. $y=\frac{4}{x}$
• D. $y=2x^2-6x+8$

Answer 1

The correct option is B $y=8x+3$

A linear equation is an equation in which the highest power of the variable is always $1$.

Option a:
$y=8x+3$
$y=8x^1+3x^0$
Here, the exponents of $x$ are $0$ and $1$. Hence, it is a linear function.

Option b: $y=2x^2-6x+8$
$y=2x^2-6x+8$ can be written as $y=2x^2-6x^1+8x^0$.
Here, the exponents of x are $0,1,$ and $2$. Hence, it is a nonlinear function.

Option c: $y=x+x^{\frac{1}{4}}$
$y=x^1+x^{\frac{1}{4}}$.
Here, the exponents of $x$ are $1$ and $\frac{1}{4}$. The exponents are not whole numbers. Hence, $y=x+x^{\frac{1}{4}}$ is a nonlinear function.

Option d: $y=\frac{4}{x}$

$\Rightarrow y=4x^{-1}$
Here, the exponent of $x$ is a negative number, $-1$. In linear functions, the exponent of $x$ is always $0$ or $1$. Therefore, it is a nonlinear function.