Question

1. $f(x)=(x + 4)^2+0$

Explanation:

Step1: Identify the vertex - form of quadratic function

The vertex - form of a quadratic function is $y=a(x - h)^2+k$, where $(h,k)$ is the vertex of the parabola. For the function $f(x)=(x + 4)^2+0$, we have $a = 1$, $h=-4$ and $k = 0$.

Step2: Find the vertex

The vertex of the parabola is $(-4,0)$.

Step3: Find the y - intercept

Set $x = 0$ in the function $f(x)=(x + 4)^2$. Then $f(0)=(0 + 4)^2=16$. So the y - intercept is $(0,16)$.

Step4: Plot points and draw the parabola

We can also find some other points. For example, when $x=-3$, $f(-3)=(-3 + 4)^2=1$; when $x=-5$, $f(-5)=(-5 + 4)^2=1$. Plot the vertex $(-4,0)$, the y - intercept $(0,16)$ and the points $(-3,1)$ and $(-5,1)$ and draw a smooth parabola opening upwards (since $a=1>0$).

Answer:

The graph is a parabola with vertex at $(-4,0)$, opening upwards and y - intercept at $(0,16)$.