91Ó°ÊÓ

Consider the quadratic function $f\left(x\right)=-x^2-4x+2$.

For this parabola, the graph will open downwards as the leading coefficient is negative.

Then finding the axis of symmetry, vertex, the intercepts, and a point with help of the axis of symmetry, will help us to plot the points of the graph and get the required graph.

On comparing the given function $f\left(x\right)=-x^2-4x+2$with the standard form $f\left(x\right)=a{x}^2+bx+c$we get

$$\begin{gathered} a=-1 \\ b=-4 \\ c=2 \end{gathered}$$

The axis of symmetry for the standard form is $x=-\frac{b}{2a}$.

So for the given function, the axis of symmetry is given as

$$\begin{gathered} x=-\frac{b}{2a} \\ x=-\frac{-4}{2·(-1)} \\ x=-2 \end{gathered}$$

Thus the equation of the axis of symmetry is $x=-2$.

Vertex is a point on the line of axis of symmetry.

Here axis of symmetry is $x=-2$.

So the function value $f(-2)$is given as

$$\begin{gathered} f(-2)=-{(-2)}^2-4·(-2)+2 \\ f(-2)=-4+8+2 \\ f(-2)=6 \end{gathered}$$

So the vertex of the parabola is $(-2,6)$.

To find the y-intercept substitute $0$for $x$in the function.

$$\begin{gathered} f\left(0\right)=-0^2-4·0+2 \\ f\left(0\right)=2 \end{gathered}$$

Thus they intercept is $(0,2)$.

To find $x$intercepts substitute $0$for $f\left(x\right)$in the function.

$\begin{array}{rcl}f\left(x\right)& =& -x^2-4x+2\\ 0& =& -x^2-4x+2\end{array}$

Using the quadratic formula we get

$$\begin{gathered} x=\frac{-(-4)Â\pm \sqrt{{(-4)}^2-4·(-1)·2}}{2·(-1)} \\ x=\frac{4Â\pm \sqrt{16+8}}{-2} \\ x=-2Â\pm \frac{2\sqrt{6}}{-2} \\ x=-2Â\pm \sqrt{6} \end{gathered}$$

On approximation we get

$$\begin{gathered} x=-2+\sqrt{6} \\ x=0.45 \end{gathered}$$

and

$$\begin{gathered} x=-2-\sqrt{6} \\ x=-4.45 \end{gathered}$$

Thus the$x$intercepts are$(0.45,0),(-4.45,0)$.

We have y intercept as $(0,2)$.

The axis of symmetry is $x=-2$.
It means that the intercept is at a distance of two units to the right of the axis of symmetry.
We can get a point on the graph using this information.
A point at the same distance, i.e., two units, but to the left will also lie on the graph. This point will be $(-4,2)$.

Plotting all the points we got and joining these points will give us the graph as-