Question

What are the coordinates of the points $y=x^2-x-12$ where it cuts the $x$ and $y$ axis?

• A. The curve cuts $x$ axis at point $(-4,0),(3,0)$ and $y$ axis at point $(0,-12)$
• B. The curve cuts $x$ axis at point $(4,0),(-3,0)$ and $y$ axis at point $(0,12)$
• C. The curve cuts $x$ axis at point $(4,0),(-3,0)$ and $y$ axis at point $(0,-12)$
• D. None of these

Answer 1

The correct option is C The curve cuts $x$ axis at point $(4,0),(-3,0)$ and $y$ axis at point $(0,-12)$

The equation $y=x^2-x-12$ cuts x axis at $y=0$
put $y=0$, we get $x^2-x-12=0$
$x^2-x-12=0$
$x^2-4x+3x-12=0$
$x(x-4)+3(x-4)=0$
$(x-4)(x+3)=0$
$x=4$ or $x=-3$
The points cut x axis are $(4,0)$ and $(-3,0)$
we put x=0 , when it cuts y axis
$y=0^2-0-12$
$y=-12$
Therefore, the required point is $(0,-12)$