Question

graph the function.
$h(x) = x^2 + 2x$

Explanation:

Step1: Analyze the function type

The function \( h(x) = x^2 + 2x \) is a quadratic function, which has the general form \( y = ax^2+bx + c \) (here \( a = 1 \), \( b = 2 \), \( c = 0 \)). The graph of a quadratic function is a parabola. For \( a>0 \) (here \( a = 1>0 \)), the parabola opens upwards.

Step2: Find the vertex (optional for graph check)

We can complete the square to find the vertex. \( h(x)=x^2 + 2x=(x + 1)^2-1 \). So the vertex is at \( (-1,-1) \)? Wait, but in the given graph, the vertex seems to be at \( (0,0) \)? Wait, maybe there is a mistake in the graph or maybe we need to check the points. Wait, let's check the point \( (5,y) \). If \( x = 5 \), \( h(5)=5^2+2\times5=25 + 10=35 \)? No, the green dot at \( x = 5 \) has \( y = 5 \)? Wait, maybe the graph is not of \( h(x)=x^2 + 2x \). Wait, maybe the function is \( h(x)=x^2 - 2x \)? No, the user provided the function as \( h(x)=x^2 + 2x \). Alternatively, maybe the graph is scaled differently. But the task is to graph the function. To graph \( h(x)=x^2 + 2x \):

• Find the x - intercepts: Set \( h(x)=0 \), \( x^2 + 2x=x(x + 2)=0 \), so \( x = 0 \) or \( x=-2 \).
• Find the vertex: The x - coordinate of the vertex of \( y = ax^2+bx + c \) is \( x=-\frac{b}{2a} \). For \( h(x) \), \( x =-\frac{2}{2\times1}=-1 \), and \( h(-1)=(-1)^2+2\times(-1)=1 - 2=-1 \), so vertex at \( (-1,-1) \).

But the given graph has a vertex at \( (0,0) \) and a point at \( (5,5) \). Maybe there is a mislabeled function. However, if we assume that the graph is supposed to represent \( h(x)=x^2 + 2x \), we can plot points:

• When \( x = 0 \), \( h(0)=0 \), so the point \( (0,0) \) is on the graph.
• When \( x = 1 \), \( h(1)=1 + 2=3 \)
• When \( x=-2 \), \( h(-2)=4-4 = 0 \)

The parabola opens upwards, passes through \( (0,0) \) and \( (-2,0) \), with vertex at \( (-1,-1) \). But the given graph has a parabola opening upwards, passing through \( (0,0) \) and a point at \( (5,5) \), which may be a different scaling or a different function, but the key is to recognize the shape of the quadratic function (upward - opening parabola) which matches the general shape of \( h(x)=x^2 + 2x \) (since it's a quadratic with \( a>0 \)).

Answer:

To graph \( h(x)=x^2 + 2x \):

1. Recognize it is an upward - opening parabola (since \( a = 1>0 \)).
2. Find x - intercepts at \( x = 0 \) and \( x=-2 \) (by solving \( x(x + 2)=0 \)).
3. Find the vertex at \( (-1,-1) \) (using \( x =-\frac{b}{2a}=-1 \) and \( h(-1)=-1 \)).
4. Plot additional points (e.g., \( x = 1 \), \( h(1)=3 \); \( x = 2 \), \( h(2)=8 \)) and draw the parabola through these points.

(Note: The given graph in the problem may have inconsistencies with the function \( h(x)=x^2 + 2x \), but the process to graph the function is as above.)