# How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of #y = x^2 + 6x + 13#?

##### 1 Answer

This is a quadratic function of form

In this case, a = 1, b = 6 and c = 13.

Since

The y-intercept value is

The x-intercepts are when y = 0 and hence are the roots of the equation

Solving this equation we get

The turning point (minimum value in this case since arms go up) is the point where the derivative of the function is zero, ie when

Hence the axis of symmetry is

The corresponding y-value is then

Hence the vertex occurs at

The domain is all the possible x-values allowed as inputs and in this case is all real numbers

The range is all possible y-values allowed as outputs and in this case is all real numbers greater than or equal to 4, that is the interval

Graphically :

graph{x^2+6x+13 [-16.82, 15.2, -0.92, 15.1]}