Question

Graph$x=-y^2-4y+12$by using properties.

Short answer

The graph for the equation $x=-y^2-4y+12$is,

Step-by-step solution

Step 1 Given equation is,

$x=-y^2-4y+12$.

Compare the given equation with $x=a{y}^2+by+c$

Identify the constants $a,b,c$.

$a=-1,b=-4,c=12$.

Since $a=-1$, the parabola opens leftward.

To find the axis of symmetry,$y=\frac{-b}{2a}$.

Substitute the known values in the formula.

$$\begin{gathered} y=\frac{-(-4)}{2\left(-1\right)} \\ =-2 \end{gathered}$$

Step 2 The vertex is on the  line y=-2.

Substitute $y=-2$in the given equation.

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

The vertex is,$\left(16,-2\right)$.

To find the $y$-intercepts substitute $x=0$in the given equation.

$$\begin{gathered} 0=-y^2-4y+12 \\ \left(y+6\right)\left(y-2\right)=0 \\ y=-6,2 \end{gathered}$$

The $y$- intercepts are,$\left(0,-6\right)$and $\left(0,2\right)$.

To Find the $x$- intercepts, substitute $y=0$in the given equation.

$$\begin{gathered} x=0-0+12 \\ x=12 \end{gathered}$$

The $x-$intercepts are, $\left(12,0\right)$.

The point $\left(12,0\right)$is two units above the line of symmetry. The symmetric point two units below the line of symmetry is $\left(12,-4\right)$.

Step 3 Plot a graph by using the known values.