Question

graph the following function by moving the green and blue dots (if necessary).
$y=(x + 5)^2-3$
start over

Explanation:

Step1: Identify the vertex form of parabola

The general vertex - form of a parabola is $y=a(x - h)^2+k$, where $(h,k)$ is the vertex. For the function $y=(x + 5)^2-3$, we have $a = 1$, $h=-5$, and $k=-3$. So the vertex of the parabola is $(-5,-3)$.

Step2: Find the y - intercept

Set $x = 0$ in the function $y=(x + 5)^2-3$. Then $y=(0 + 5)^2-3=25-3=22$. So the y - intercept is the point $(0,22)$.

Step3: Find the x - intercepts

Set $y = 0$ in the function $y=(x + 5)^2-3$. Then $(x + 5)^2=3$, so $x+5=\pm\sqrt{3}$, and $x=-5\pm\sqrt{3}$. The x - intercepts are approximately $x=-5+\sqrt{3}\approx - 3.27$ and $x=-5-\sqrt{3}\approx - 6.73$.

Step4: Analyze the shape

Since $a = 1>0$, the parabola opens upward.

To graph the function, move the green and blue dots to the vertex $(-5,-3)$, y - intercept $(0,22)$ and the x - intercepts $(-5+\sqrt{3},0)$ and $(-5 - \sqrt{3},0)$ and draw a smooth parabola opening upward passing through these points.

Answer:

Graph the parabola with vertex at $(-5,-3)$, y - intercept at $(0,22)$ and x - intercepts at $(-5+\sqrt{3},0)$ and $(-5-\sqrt{3},0)$ opening upward.