Question

graph the quadratic function f(x)=3(x + 1)^2-3 and label the points on the image.

Explanation:

Step1: Identify the vertex form

The quadratic function is in vertex - form $y = a(x - h)^2+k$, where $(h,k)$ is the vertex. For $f(x)=3(x + 1)^2-3$, $a = 3$, $h=-1$, $k = - 3$. So the vertex is $(-1,-3)$.

Step2: Find the y - intercept

Set $x = 0$ in $f(x)=3(x + 1)^2-3$. Then $f(0)=3(0 + 1)^2-3=3\times1 - 3=0$. So the y - intercept is $(0,0)$.

Step3: Find the x - intercepts

Set $y = 0$ in $f(x)=3(x + 1)^2-3$. Then $3(x + 1)^2-3=0$. First, add 3 to both sides: $3(x + 1)^2=3$. Divide both sides by 3: $(x + 1)^2=1$. Take the square - root of both sides: $x+1=\pm1$. Solve for $x$: $x=-1\pm1$. So $x = 0$ or $x=-2$. The x - intercepts are $(0,0)$ and $(-2,0)$.

Step4: Analyze the shape

Since $a = 3>0$, the parabola opens upward.

Answer:

The vertex of the parabola is $(-1,-3)$, the y - intercept is $(0,0)$, and the x - intercepts are $(0,0)$ and $(-2,0)$. The parabola opens upward. To graph, plot the vertex, the x - intercepts, and the y - intercept, and draw a smooth upward - opening curve through these points.