Find the vertex and axis of symmetry, and intercepts for a quadratic equation y= x² + 6x + 5?

Solution:

Given function f(x) = x² + 6x + 5

The axis of symmetry is the x-coordinate of the vertex, a vertical line across which the graph exhibits symmetry, given by x = -b/2a when the quadratic is in the form ax² + bc + c

Here, we see b = 6,a = 1; thus, the axis is x = -6/2 = -3

The coordinates of the vertex are given by (-b/2a, f(-b/2a))

We know -b/2a = -3, so we need f(-3).

f(-3) = (-3)(-3) + 6(-3) +5 = 14 - 18 = -4

The vertex is then (-3, -4)

X-intercept is f(x) when y=0 : x² + 6x + 5 = 0 : x = -1,-5

X-intercepts are (-1, 0) and (-5, 0)

Y-intercepts is f(x) when x = 0: y = 5

Y-intercepts are (0, 5)