Question

solve the equation by graphing.$-x^{2}-2x-6=0$first, graph the associated parabola by plotting the vertex and four additional points, two on each side of thethen, 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: Rewrite in standard parabola form

Rewrite $-x^2 - 2x - 6 = 0$ as the function $y = -x^2 - 2x - 6$. This is a downward-opening parabola (since the coefficient of $x^2$ is negative).

Step2: Find the vertex

Use the vertex formula for $ax^2+bx+c$: $x = -\frac{b}{2a}$. Here $a=-1$, $b=-2$.
$x = -\frac{-2}{2(-1)} = \frac{2}{-2} = -1$
Substitute $x=-1$ into the function to find $y$:
$y = -(-1)^2 -2(-1) -6 = -1 +2 -6 = -5$
Vertex: $(-1, -5)$

Step3: Find points left of vertex

Choose $x=-2$:
$y = -(-2)^2 -2(-2) -6 = -4 +4 -6 = -6$ → Point: $(-2, -6)$
Choose $x=-3$:
$y = -(-3)^2 -2(-3) -6 = -9 +6 -6 = -9$ → Point: $(-3, -9)$

Step4: Find points right of vertex

Choose $x=0$:
$y = -(0)^2 -2(0) -6 = 0 +0 -6 = -6$ → Point: $(0, -6)$
Choose $x=1$:
$y = -(1)^2 -2(1) -6 = -1 -2 -6 = -9$ → Point: $(1, -9)$

Step5: Analyze x-intercepts

The equation $-x^2 -2x -6=0$ corresponds to finding where $y=0$ (x-intercepts). The vertex is at $(-1, -5)$, and the parabola opens downward, so all points on the parabola have $y \leq -5$. The parabola never crosses the x-axis ($y=0$).

Answer:

No solution