Answer
See below:
Work Step by Step
Consider the quadratic function,
\[f\left( x \right)={{x}^{2}}-2x-8\]
As the quadratic function also known as a parabola.
The general form of the quadratic function is as follows:
\[f\left( x \right)=a{{x}^{2}}+bx+c\]
In which \[a,b,c\] are real numbers and \[a\ne 0\].
Compare given quadratic function with the general form of quadratic function, then the resulting value will be written below:
\[a=1,b=\left( -2 \right),c=\left( -8 \right)\]
To draw the graph of the quadratic equation, there are various steps that must follow:
Step (1): See the coefficient of \[{{x}^{2}}\] if it is positive, then parabola is open upwards and if it is negative, then parabola is open downwards.
In the given function coefficient of \[{{x}^{2}}\] is positive so parabola opens upwards.
Step (2): Obtain the vertex of the parabola, which is\[x=\frac{\left( -b \right)}{2a}\].
In given function vertex of the parabola is written below:
\[x=\frac{\left( -b \right)}{2a}\]
Substitute the value of \[b\ \text{and}\ a\]in the above formulae which are:
\[\begin{align}
& x=\frac{\left( -b \right)}{2a} \\
& =\frac{\left( -\left( -2 \right) \right)}{2\left( 1 \right)} \\
& =\frac{2}{2} \\
& =1
\end{align}\]
So, vertex of given parabola is \[1\]
Step (3): Substitute \[x\]coordinate value in parabola equation to find \[y\]coordinate.
\[\begin{align}
& y={{x}^{2}}-2x-8 \\
& ={{1}^{2}}-2\left( 1 \right)-8 \\
& =1-2-8 \\
& =-9
\end{align}\]
So, the vertex of the parabola is\[\left( 1,-9 \right)\].
Step (4): Obtain the \[x\]intercepts by substitute \[y=0\], then the quadratic equation will be formed.
Substitute \[y=0\] for the given quadratic function which is:
\[\begin{align}
& y={{x}^{2}}-2x-8 \\
& 0={{x}^{2}}-2x-8 \\
& {{x}^{2}}-2x-8=0
\end{align}\]
Further, solve \[x\] by quadratic formulae which are:
\[x=\frac{-b\pm \sqrt{{{b}^{2}}-4\cdot a\cdot c}}{2a}\]
Then,
\[\begin{align}
& x=\frac{-b\pm \sqrt{{{b}^{2}}-4\cdot a\cdot c}}{2a} \\
& =\frac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\cdot 1\cdot \left( -8 \right)}}{2\cdot 1} \\
& =\frac{2\pm \sqrt{4+32}}{2} \\
& =\frac{2\pm \sqrt{36}}{2}
\end{align}\]
Further, solve:
\[\begin{align}
& x=\frac{2\pm \sqrt{36}}{2} \\
& =\frac{2\pm 6}{2}
\end{align}\]
First, take positive sign,
\[\begin{align}
& x=\frac{2+6}{2} \\
& =\frac{8}{2} \\
& =4
\end{align}\]
Then, take negative sign,
\[\begin{align}
& x=\frac{2-6}{2} \\
& =\frac{-4}{2} \\
& =-2
\end{align}\]
Here getting the two values of \[x\], it shows that the graph cut the \[x\]axis two times which is \[\left( -2,0 \right)\ \text{and}\ \left( 4,0 \right)\]
Step (5): Obtain the \[y\]intercepts by substituting \[x=0\] , then the equation will be formed as:
\[\begin{align}
& y={{x}^{2}}-2x-8 \\
& =0-2\left( 0 \right)-8 \\
& =-8
\end{align}\]
Basically, the \[y\]intercept is \[c\] and the parabola passes through \[\left( 0,-8 \right)\].
Step (6): Connect all the points with a curve.
The graph of the parabola is drawn below:
