Sketch it's graph

The equation is $y = -x^2 - 2x + 5$. Since the coefficient of $x^2$ is $-1$ (which is negative), the parabola opens downwards.

The x-coordinate of the vertex for a parabola $y = ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$.
For $y = -x^2 - 2x + 5$, we have $a=-1$ and $b=-2$.

Now, substitute $x=-1$ into the equation to find the y-coordinate:

The vertex is at $(-1, 6)$.

Set $x=0$ in the equation:

The y-intercept is at $(0, 5)$.

Set $y=0$ in the equation:

Rearrange to $x^2 + 2x - 5 = 0$.
Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

The x-intercepts are approximately $x_1 = -1 + \sqrt{6} \approx -1 + 2.45 = 1.45$ and $x_2 = -1 - \sqrt{6} \approx -1 - 2.45 = -3.45$.
So, the x-intercepts are approximately $(1.45, 0)$ and $(-3.45, 0)$.

Plot the vertex $(-1, 6)$, the y-intercept $(0, 5)$, and the x-intercepts $(-1+\sqrt{6}, 0)$ and $(-1-\sqrt{6}, 0)$. Draw a smooth parabolic curve opening downwards through these points.

Summary of key points for sketching:

• Opens downwards.
• Vertex: $(-1, 6)$
• Y-intercept: $(0, 5)$
• X-intercepts: $(-1+\sqrt{6}, 0)$ and $(-1-\sqrt{6}, 0)$ (approximately $(1.45, 0)$ and $(-3.45, 0)$)