h(x)=x^2+8x

The function you've provided is a quadratic function, \( h(x) = x^2 + 8x \). This can be rewritten in standard form or vertex form if needed.

1. Standard Form

The standard form of a quadratic function is already given as \( h(x) = ax^2 + bx + c \), where \( a = 1 \), \( b = 8 \), and \( c = 0 \).

2. Vertex Form

To convert this into vertex form, we can complete the square:

1. Start with \( h(x) = x^2 + 8x \).

2. Take the coefficient of \( x \) (which is 8), halve it to get 4, and then square it to get 16.

3. Rewrite the function by adding and subtracting 16:

   \[ h(x) = (x^2 + 8x + 16) - 16 \]

4. This simplifies to:

   \[ h(x) = (x + 4)^2 - 16 \]

Now, the vertex form of the function is \( h(x) = (x + 4)^2 - 16 \). This tells us that the vertex of the parabola is at the point (-4, -16).

3. Finding the Vertex

• The vertex from the vertex form is \( (-4, -16) \).

4. Axis of Symmetry

• The axis of symmetry for the parabola is the vertical line \( x = -4 \).

5. Intercepts

• y-intercept: Set \( x = 0 \):

  \[ h(0) = 0^2 + 8(0) = 0 \]

So the y-intercept is \( (0, 0) \).

• x-intercepts: Set \( h(x) = 0 \):

  \[ x^2 + 8x = 0 \]

  Factoring gives:

  \[ x(x + 8) = 0 \]

Thus, \( x = 0 \) or \( x = -8 \) gives x-intercepts at \( (0, 0) \) and \( (-8, 0) \).

Summary

• The function is \( h(x) = x^2 + 8x \).
• Vertex: \( (-4, -16) \)
• Axis of symmetry: \( x = -4 \)
• y-intercept: \( (0, 0) \)
• x-intercepts: \( (0, 0) \) and \( (-8, 0) \)

If you have any specific questions or need further calculations, feel free to ask!