Draw the graph of \[y={{x}^{2}}-x-6\] and find the zeros. Justify the answer.
VerifiedComplete step by step solution:
First we will rearrange the equation.
Equation is \[y={{x}^{2}}-x-6\]
Therefore,
\[y={{x}^{2}}-x-6\]
\[y={{x}^{2}}-\dfrac{2\times 1}{2}x-6-{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{1}{2} \right)}^{2}}\]
To re-arrange in the form of a quadratic equation.
\[ y=\left( {{x}^{2}}-\dfrac{2\times 1}{2}x+{{\left( \dfrac{1}{2} \right)}^{2}} \right)-6-{{\left( \dfrac{1}{2} \right)}^{2}}\]
To get \[{{\text{a}}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}\]
\[ y={{\left( x-\dfrac{1}{2} \right)}^{2}}-\dfrac{25}{4} \]
\[ y+\dfrac{25}{4}={{\left( x-\dfrac{1}{2} \right)}^{2}} \]
Now since we have an equation which is very much similar to the general equation of a parabola we can compare it to
\[ (x-\alpha)^2= 4a (y- \beta) \]
where $\alpha ,\beta $ are the vertex from where a parabola starts.
And a is the axis which is the distance between vertex and the focus in x-axis.
After comparing it to the equation of parabola with directrix in y-axis we get,
\[\beta =-\dfrac{25}{4}\] and \[ \alpha =\dfrac{1}{2}\]
$\therefore \left( +\dfrac{1}{2},-\dfrac{25}{4} \right)$ is the vertex of the parabola. Therefore the graph of this equation will be:
The given equation was matching with \[{{(x-\alpha )}^{2}}=4a(y-\beta )\text{ }\]which has the directrix in the y-axis. Moreover, vertex was \[\left( +\dfrac{1}{2},-\dfrac{25}{4} \right)\] or \[\left( +0.5,-6.25 \right)\] thus the parabola stars from the 4th quadrant.
The next step will be to put \[y=0\] to find the value where the equation is 0. That also signifies that our value of y will be 0, since \[y={{x}^{2}}-x-6\];
\[ \therefore {{x}^{2}}-3x+2x-6=0 \]
\[x(x-3)+2(x-3)=0 \]
\[ (x-3)(x+2)=0 \]
\[ \therefore x=-2,3 \]
So we get the value where \[y=0\text{ and }x=-2,3\].
Thus the coordinates will be \[A(-2,0)\text{ and }B(3,0)\],
Now, to further verify these values and coordinates we will put the value of x we got from coordinate A and B in the equation \[y={{x}^{2}}-x-6\],
\[ \therefore y={{(-2)}^{2}}-(-2)-6= 4+2-6 =0 \]
& and \[y={{(3)}^{2}}-(3)-6= 9-3-6 =0 \]
And since, we can see that the values of x which we got from solving the equation by putting \[y=0\] , we can say that the values are accurate. Thus the graph shown above is also accurate.
The \[coordinate\,D(0,-6)\] is not taken into account or notice is because we are trying to find where \[y=0\]. We are not finding where the value of \[x=0\]. Thus the \[coordinate\,D(0,-6)\]is taken off from notice. But the fact remains that this coordinate also has a zero so to find it we can simply put \[x=0\text{ in }y{{=}^{2}}-x-6\,\text{ which will give us }y=-6.\]thus giving us the coordinate D which is \[D(0,-4)\].
Note:The value of \[\text{ }\!\!'\!\!\text{ }a'\text{ }\]which we found out earlier is not required in this question which is why it is ignored. \[\left( \dfrac{1}{2}=0.5\text{ } \right)\text{and }\left( -\dfrac{25}{4}=-6.25 \right)\]
