91Ó°ÊÓ

We are given a function$f\left(x\right)=x^2+4x−12$.

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,

$x=\frac{-b}{2a}$

putting $$\begin{gathered} a=1 \\ b=4 \end{gathered}$$, we get

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

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,

$$\begin{gathered} f(-2)=(-2{)}^2+4(-2)-12 \\ f(-2)=4-8-12 \\ f(-2)=-16 \end{gathered}$$

Hence the vertex is $(-2,-16)$.

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

role="math" localid="1645165222730" $$\begin{gathered} f\left(0\right)=(0{)}^2+4(0)-12 \\ f(0)=0+0-12 \\ f(0)=-12 \end{gathered}$$

Hence they-intercept is $(0,-12)$.

They-intercept$(0,-12)$is$2$units to the right of axis of symmetry, so there is a points $2$units to left of symmetry.

The point is $(-4,-12)$.

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

$$\begin{gathered} x^2+4x-12=0 \\ {x}^2+6x-2x-12=0 \\ (x+6)(x-2)=0 \\ x+6=0 \\ x=-6 \end{gathered}$$

or

$$\begin{gathered} x-2=0 \\ x=2 \end{gathered}$$

So, the $x-$intercepts are $-6,2$.

The graph is as follows,