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

Sep 26, 2015

This is a quadratic function of form $y = a {x}^{2} + b x + c$.
In this case, a = 1, b = 6 and c = 13.

Since $a > 0 \implies$ the arms go up.

The y-intercept value is $c = 13$

The x-intercepts are when y = 0 and hence are the roots of the equation ${x}^{2} + 6 x + 13 = 0$
Solving this equation we get
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = - 3 \pm 2 i \in \mathbb{C}$, and so there is no x-intercepts.

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
$2 x + 6 = 0 \implies x = - 3$
Hence the axis of symmetry is $x = - 3$
The corresponding y-value is then $y \left(- 3\right) = 4$
Hence the vertex occurs at (-3;4)

The domain is all the possible x-values allowed as inputs and in this case is all real numbers $\mathbb{R}$
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 [4;oo)

Graphically :

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