91Ó°ÊÓ

We are given a function$f\left(x\right)=x^2+4x+3$.

The parabola will open upwards here, because the leading coefficient is positive. Then by finding the axis of symmetry, vertex, the intercepts and a point with help of axis of symmetry, will help us to plot the points of the graph and get the required graph.

The axis of symmetry is given by,

Putting $a=1$a and$b=4$, we get

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

The equation of axis of symmetry is $x=-2$.

Vertex is a point on the line of axis of symmetry. Here axis of symmetry is $x=-2$.

We will get the vertex by putting $x=-2$in the given function,

$$\begin{gathered} f(-2)=(-2{)}^2+4(-2)+3 \\ f(-2)=4-8+3 \\ f(-2)=-1 \end{gathered}$$

Hence the vertex is $(-2,-1)$.

We get y-intercept by putting $x=0$,

$$\begin{gathered} f\left(0\right)=0^2+4\left(0\right)+3 \\ f\left(0\right)=3 \end{gathered}$$

Hence they-intercept is $(0,3)$.

The y-intercept$(0,3)$is $2$units to the right of axis of symmetry, so there is a points $2$units to left of symmetry.

The point is $(-4,3)$.

The$x-$intercept is given by putting the function equal to zero, so

$$\begin{gathered} x^2+4x+3=0 \\ {x}^2+3x+x+3=0 \\ (x+1)(x+3)=0 \\ x+1=0 \\ x=-1 \end{gathered}$$

or

$$\begin{gathered} x+3=0 \\ x=-3 \end{gathered}$$

So, the $x-$intercepts are $-1,-3$.

The graph is as follows,