二次函数零点及抛物线顶点求解咨询：g(x)=4x²+36

Hey James, let's break down this problem step by step—you’ve already made solid progress by factoring out the 4, so let’s build on that!

Finding the Zeros (Roots) #

To find where the function crosses the x-axis (zeros), we set g(x) = 0 and solve for x. Starting with your factored form:

4(x² + 9) = 0

Since 4 is a non-zero constant, we can divide both sides by 4 to simplify the equation to:

x² + 9 = 0

Rearranging gives x² = -9.

• In the real number system, there are no solutions here—you can’t square a real number to get a negative result. That means this parabola never touches or crosses the x-axis.
• If you’re working with complex numbers, the zeros would be x = 3i and x = -3i (where i is the imaginary unit, defined as i² = -1).

Finding the Parabola’s Vertex #

There are two straightforward ways to find the vertex of this quadratic:

1. Vertex Form Method
   Rewrite the function in the vertex form a(x - h)² + k, where (h, k) is the vertex. Your function is already in a form we can use:

   g(x) = 4x² + 36 = 4(x - 0)² + 36
   

   Here, h = 0 and k = 36, so the vertex is at the point (0, 36).

2. Standard Formula Method
   For a quadratic in standard form ax² + bx + c, the x-coordinate of the vertex is calculated using x = -b/(2a). In g(x) = 4x² + 0x + 36, we have a = 4 and b = 0. Plugging these values in:

   x = -0/(2*4) = 0
   

   Now substitute x = 0 back into the function to find the y-coordinate:

   g(0) = 4*(0)² + 36 = 36
   

   This confirms the vertex is at (0, 36).

内容的提问来源于stack exchange，提问作者James Leaf