Question

The graph of polynomial $P\left(x\right)=ax-b$, where $a\ne 0;a,b\epsilon R$ intersect intersects $X-axis$ at

• A. $(\frac{b}{a},0)$
• B. $(0,\frac{b}{a})$
• C. $(-\frac{b}{a},0)$
• D. $(-\frac{a}{b},0)$

Answer 1

The correct option is A $(\frac{b}{a},0)$

The graph of the polynomial $p\left(x\right)=ax+b;a\ne 0,a,b\epsilon R$

contains at least two distinct points and it intersects X-axis.

Let $p\left(x\right)=0⟹x=-\frac{b}{a}$

So, it intersects X-axis at exactly one point $(-\frac{b}{a},0)$

where $\frac{b}{a}$ is the zero of $p\left(x\right)$.

Now, the polynomial $P\left(x\right)=ax-b$ can be written as $P\left(x\right)=ax+(-b)$

Hence, it intersects X-axis at $(\frac{b}{a},0)$