Question

solve the equation by graphing.$x^{2}+12x+36=0$first, graph the associated parabola by plotting the vertex and four additional points, two on eachthen, use the graph to give the solution(s) to the equation.if there is more than one solution, separate them with commas.if there is no solution, click
o solution\.progress: 0/2

Explanation:

Step1: Find vertex of parabola

For quadratic $ax^2+bx+c$, vertex $x=-\frac{b}{2a}$.
Here $a=1, b=12$, so $x=-\frac{12}{2(1)}=-6$.
Substitute $x=-6$: $y=(-6)^2+12(-6)+36=36-72+36=0$.
Vertex: $(-6, 0)$

Step2: Find points left of vertex

Choose $x=-7, x=-8$:
For $x=-7$: $y=(-7)^2+12(-7)+36=49-84+36=1$ → $(-7,1)$
For $x=-8$: $y=(-8)^2+12(-8)+36=64-96+36=4$ → $(-8,4)$

Step3: Find points right of vertex

Choose $x=-5, x=-4$:
For $x=-5$: $y=(-5)^2+12(-5)+36=25-60+36=1$ → $(-5,1)$
For $x=-4$: $y=(-4)^2+12(-4)+36=16-48+36=4$ → $(-4,4)$

Step4: Identify x-intercepts (solutions)

The parabola touches the x-axis only at $x=-6$.

Answer:

$-6$