Question

What is the axis of symmetry of h(x)=5x2+40x+64

Answer 1

The axis of symmetry of a parabola is a vertical line that divides the parabola into two similar halves. The axis of symmetry always passes through the vertex of the parabola . The x -coordinate of the vertex is the equation of the axis of symmetry of the parabola. For the function h(x) = 5x^2 + 40x + 64 we write it in vertex form as follows: h(x) = 5(x^2 + 8x) + 64 h(x) = 5(x^2 + 8x + 16) + 64 - 5(16) h(x) = 5(x + 4)^2 + 64 - 80 h(x) = 5(x - (-4))^2 - 16 Thus the vertex is given by (h, k) = (-4, -16) Therefore, the axis of symmetry of h(x) = 5x^2 + 40x + 64 is the line x = -4

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Answer 2

Answer:

x = -4.

Step-by-step explanation:

As we have the standard form of a parabola, that is $ax^2+bx+c$ the axis of symetry will be the line $x = \frac{-b}{2a}$. In this case we have that a = 5, b = 40 and c = 64. Then the axis of symtery is

$x = \frac{-40}{10} = -4$.